L 



li— MB 



IHIIII 



I 



JVVHWhU 

HHiiU 



r 



■NffinJIilllHl 



m 
KM 






MHHHIHIl MWyHHU 

w 



mm 



JMM 



n 



11 
li 



"'IHillfl HI 



I 



9H 






/ 




Class JL 



Book. 



(topyrigttW 



COPYRIGHT DEPOSKR 



DIFFERENTIAL AND INTEGRAL CALCULUS 



DIFFERENTIAL AND INTEGRAL 
CALCULUS 



AN INTRODUCTORY COURSE 
FOR 

COLLEGES AND ENGINEERING SCHOOLS 



BY 

LORRAIN S. HULBURT 

it 

COLLEGIATE PROFESSOR OP MATHEMATICS 
IN THE JOHNS HOPKINS UNIVERSITY 



LONGMANS, GREEN, AND CO 

FOURTH AVENUE & 30TH STREET, NEW YORK 

LONDON, BOMBAY AND CALCUTTA 

1912 






Copyright, 1912, by 
LONGMANS, GREEN, AND CO. 



Stanbope flbress 

H.GILSON COMPANY 
BOSTON, U.S.A. 



; 

gd.A3l4949 



PREFACE 

This book is designed as an elementary textbook of the Cal- 
culus. For many years the author has taught this subject with- 
out a textbook, using manifolded lecture notes of his own placed 
in the hands of his students. These notes have been rewritten 
from time to time, and this book is the result. 

In the frequent revision of his lecture notes the author has 
made free use of the current textbooks of Calculus. Conse- 
quently, while many of the problems and exercises in this book 
have become, from long use, the common property of all mathe- 
maticians, and many others are the author's own, there is doubt- 
less a residue derived from other sources. And now, after the 
lapse of years, the author is for the most part unable to say just 
what these sources are. Of the older textbooks those of Byerly 
and Osborne are probably the ones that have contributed most 
largely to the lists of exercises. And the author is certainly in- 
debted to the more recent textbook of Osgood, from which he 
has derived helpful suggestions. Also a number of the curves 
are taken from Loria-Schtitte's Ebene Kurven. 

The author desires to express here his gratitude to his friends 
for their cordial encouragement and aid. In particular his thanks 
are due to his colleague, Dr. A. Cohen, for valuable suggestions 
in writing the chapters on Differential Equations; to Professor 
H. S. White of Vassar College, for helpful criticisms and sugges- 
tions, and for aid in the proof reading; to Mr. D. M. Liddell of 
the Engineering and Mining Journal, who has read critically 
all the proof sheets; and to Dr. J. R. Conner of The Johns 
Hopkins University, who has verified the answers to many of the 
exercises. 

L. S. HULBURT. 

The Johns Hopkins University, 
Baltimore, Md., April, 1912. 



CONTENTS 

BOOK I 

DIFFERENTIAL CALCULUS 

CHAPTER I 

INTRODUCTION 

Article Page 

1. Variables and Constants 1 

2. Functions 1 

3. Functional Symbols 4 

4. The Identity 5 

5. Exercises 6 

6. Kinds of Functions 7 

7. Graphs of Functions 9 

8. Discontinuities of Functions 10 

CHAPTER II 

LIMITS 

9. Definition of Limit 16 

10. Infinitesimals and Infinities 18 

11. Limits of Functions 20 

12. The Difference between the Limit of/ (x) when x = a and the Value 

of / (x) when x = a 22 

13. Further Examples of Finding Limits of Functions 23 

14. Exercises 24 

15. Meaning of the Symbols t; = °° and ^- = 24 

16. The Limit of a Rational Fraction in x, as x = oo 25 

17. Infinite Limits of Trigonometric Functions 26 

18. Example of a Function that has no Limit 27 

19. A Geometrical Limit 28 

20. General Theorems of Limits 28 

CHAPTER III 

THE DERIVATIVE 

21. Increments 32 

22. The Derivative 33 

vii 



Vlll CONTENTS 

Article Page 

23. The Equations of Tangent and Normal 36 

24. Exercises 37 

CHAPTER IV 
RULES FOR DIFFERENTIATION 

25. Definitions 38 

26. Differentiation of Polynomials 38 

27. Exercises 43 

28. Tangents and Normals of Curves 43 

29. Exercises 44 

30. Differentiation of Fractions 44 

31. Proof of VI when n is a Negative Integer 45 

32. Exercises 46 

33. Differentiation of Irrational Functions 46 

34. Exercises 48 

35. A More Descriptive Notation for the Derivative 48 

36. Exercises 49 

37. Differentiation of Implicit Functions . . '. 49 

38. Exercises : 50 

39. Differentiation of Trigonometric Functions 51 

40. Exercises 53 

41. Differentiation of Circular Functions 53 

42. Exercises 55 

CHAPTER V 
DERIVATIVES OF HIGHER ORDERS 

43. Definitions and Notation 56 

44. Successive Differentiation of Explicit Functions 56 

45. Successive Differentiation of Implicit Functions 57 

46. Exercises 58 

CHAPTER VI 

SOME GENERAL PROPERTIES OF FUNCTIONS. CONVEXITY, 
CONCAVITY, AND FLEXES 

47. The Interval 60 

48. Continuity 60 

49. Discontinuities of f(x) 63 

50. How a Function Changes Sign 63 

51. Roots of Polynomials 65 

52. Zeros of Functions 67 

53. Exercises 68 



CONTENTS ix 

Article Page 

54. Increasing and Decreasing Functions 68 

55. Convexity and Concavity 70 

56. The Flex 72 

57. Examples of Finding Flexes and Convex and Concave Arcs 73 

58. Exercises 75 

CHAPTER VII 
MAXIMA AND MINIMA 

59. Definitions 76 

60. Maxima and Minima Determined by Means of the First Derivative.. 77 

61. Exercises 78 

62. Employment of the Second Derivative in Determining Maxima 

and Minima : 79 

63. Exercises 80 

64. Curve Tracing 81 

65. Exercises 83 

66. Problems in Maxima and Minima 83 

67. Exercises 84 

CHAPTER VIII 

DIFFERENTIATION OF LOGARITHMIC AND EXPONENTIAL 
FUNCTIONS. DIFFERENTIALS 

68. To Determine ^^fl + ^ J 88 

69. Differentiation of Logarithmic Functions 90 

70. Differentiation of Exponential Functions 92 

71. Exercises 94 

72. The Compound Interest Law 94 

73. Differentials 97 

74. Differentials of Higher Orders 101 

75. Remarks on the Differential Notation 101 

CHAPTER IX 

THE DERIVATIVE AS VELOCITY 

76. Definition of Velocity 103 

77. Instantaneous Velocity 104 

78. Exercises * 107 

79. Angular Velocity 108 

80. Velocity of any Change of State 109 

81. Expansion of a Metal Rod Ill 

82. Acceleration Ill 

83. Exercises 112 

84. A General Differentiation Formula 113 



X CONTENTS 

CHAPTER X 

ADDITIONAL EXAMPLES IN CURVE TRACING 

Article Page 

85. Multiple Points 115 

86. Examples 115 

87. Exercises 117 

CHAPTER XI 

CURVES GIVEN BY PARAMETRIC EQUATIONS 

88. Parametric Equations 120 

89. Example. The Folium of Descartes ■ 120 

90. Geometric Interpretation of the Parameter 123 

91. Exercises 124 

92. Derivation of Parametric Equations 126 

CHAPTER XII 
CYCLOIDAL CURVES 

93. The Cycloid 128 

94. The Epi- and Hypo-Cycloids 129 

95. Exercises 131 

96. The Involute of the Circle 132 

CHAPTER XIII 
CURVES GIVEN BY POLAR EQUATIONS 

97. The Tangent in Polar Coordinates 134 

98. Curves in Polar Coordinates 135 

99. Exercises 137 

CHAPTER XIV 

THE DERIVATIVE OF THE ARC. TJIE METHOD OF 
INFINITESIMALS 

100. A Theorem of Geometry 140 

101. The Derivative of the Arc: Cartesian Coordinates 141 

102. The Derivative of the Arc: Polar Coordinates 143 

103. The Method of Infinitesimals 144 

104. Exercises 148 



CONTENTS XI 

CHAPTER XV 

SIMPLE FORMULAE OF KINEMATICS 

Article Page 

105. Resolution of Velocities and Accelerations 149 

106. Resolution of Velocity along a Curve 150 

107. Exercises 152 

CHAPTER XVI 
CURVATURE. EVOLUTES AND INVOLUTES 

108. Curvature 153 

109. The Circle of Curvature 155 

110. Evolutes and Involutes 157 

111. Properties of the Evolute 158 

112. Exercises 161 

CHAPTER XVII 
THE LAW OF THE MEAN. INDETERMINATE FORMS 

113. Rolle's Theorem 163 

114. The Law of the Mean 164 

115. Cauchy's Theorem 165 

116. Indeterminate Forms 166 

117. Exercises 171 

118. Indeterminate Forms (Continued) 172 

119. Exercises 173 

BOOK II 

INTEGRAL CALCULUS 

CHAPTER XVIII 
FIRST PRINCIPLES OF INTEGRATION 

120. Definitions 175 

121. Two General Theorems of Integration 176 

122. Fundamental Formulae of Integration 178 

123. Proof and Application of Formulas I and II 180 

124. Devices for Bringing the Integrand to a Form to which the Funda- 

mental Formulae Apply 181 

125. Exercises 183 

126. Proof and Application of Formulae III, IV, and V 184 

127. Additional Devices 185 

128. Exercises 186 



Xll CONTENTS 

Article Page 

129. Proof and Application of Formulae VI and VII 187 

130. Exercises 187 

131. Proof and Application of the Trigonometric Formulae VIII ... XV 187 

132. Exercises ' 188 

133. Miscellaneous Exercises in Integration 188 

CHAPTER XIX 

INTEGRATION BY AID OF A TABLE OF INTEGRALS. SOME 
SPECIAL METHODS OF INTEGRATION 

134. Explanation of the Table 190 

135. Exercises 192 

136. Some Special Methods of Integration 193 

137. Integration by Algebraic Substitution 193 

138. Exercises 194 

139. Integration of cos m x sin 71 x 195 

140. Exercises 196 

141. Integration by Trigonometric Substitution 196 

142. Exercises 197 

143. Integration by Parts 198 

144. Exercises 200 

CHAPTER XX 
SIMPLE APPLICATIONS OF INTEGRATION IN GEOMETRY 

145. Illustrative Problem 201 

146. Exercises 204 

CHAPTER XXI 
APPLICATIONS OF INTEGRATION IN KINEMATICS 

147. To Determine the Motion of a Body when the Velocity is Given. . 205 

148. Exercises 208 

149. To Determine the Motion of a Body when the Acceleration is 

Given 209 

150. Exercises 212 

151. The Motion of a Body Falling to the Earth from a Great Distance 

above the Surface 213 

152. Exercises 215 

153. The Motion of a Projectile 215 

154. Exercises 220 

155. Problem 3 221 

156. Exercises 223 



CONTENTS X1U 

CHAPTER XXII 
THE DEFINITE INTEGRAL. AREAS AND LENGTHS OF CURVES 

Article Page 

157. The Definite Integral 224 

158. Exercises 225 

159. The Derivative of Area: Cartesian Coordinates 225 

160. The Area under a Curve Expressed as a Definite Integral 226 

161. Exercises 230 

162. Areas of Curves Given by Parametric Equations 231 

163. Exercises 233 

164. The Derivative of Area : Polar Coordinates 234 

165. Areas: Polar Coordinates 235 

166. Exercises 235 

167. Lengths of Curves 236 

168. Exercises 238 

CHAPTER XXIII 

THE DEFINITE INTEGRAL AS THE LIMIT OF A SUM OF 
INFINITESIMAL PRODUCTS 

169. The Fundamental Theorem of the Integral Calculus 240 

170. Area of a Sector 244 

171. Proof that Every Function has an Integral 246 

172. The Problem of Motion 246 

173. A Theorem of Infinitesimals 248 

CHAPTER XXIV 
SURFACES AND SOLIDS OF REVOLUTION 

174. Definitions 253 

175. Volumes of Solids of Revolution 253 

176. Areas of Surfaces of Revolution 255 

177. Exercises 260 

178. Solids and Surfaces of Revolution: Polar Coordinates , 262 

179. Exercises 264 

BOOK III 

INTRODUCTION TO ANALYTIC GEOMETRY OF 
THREE DIMENSIONS 

CHAPTER XXV 
THE POINT, THE PLANE, AND THE SURFACE IN SPACE 

180. Rectangular Space Coordinates 265 

181. Relations between Two Points 266 



XIV CONTENTS 

Article Page 

182. Some Simple Loci 267 

183. The Plane in Space 268 

184. Exercises 270 

185. Quadric Surfaces 271 

186. Exercises 276 

CHAPTER XXVI 
THE RIGHT LINE IN SPACE 

187. Projection 278 

188. Direction Cosines 279 

189. Parametric Equations of the Line 281 

190. The Normal Equation of the Plane 282 

191. Exercises 283 

CHAPTER XXVII 

VOLUMES OF SOLIDS. AREAS OF CYLINDRICAL SURFACES 

192. Volumes of Solids . 286 

193. Exercises 289 

194. The Line Integral. The Area of a Cylindrical Surface 289 

195. Exercises 291 



BOOK IV 

FUNCTIONS OF MORE THAN ONE ARGUMENT 

CHAPTER XXVIII 
PARTIAL AND TOTAL DERIVATIVES 

196. Definitions 293 

197. Partial Derivatives 294 

198. Exercises 295 

199. The Tangent Plane and the Normal Line 296 

200. Exercises 299 

201. Partial Derivatives of Higher Orders 299 

202. Exercises 301 

203. The Total Derivative 301 

204. Exercises : 306 

205. Partial Derivatives of / (x, y, z, . . . . ) where x,y,z, are them- 

selves Functions of Several Independent Arguments 307 

206. Exercises 310 

207. Implicit Functions 311 



CONTENTS XV 

Article Page 

208. Exercises 313 

209. The Tangent Plane and the Normal Line 314 

210. Differentials, Partial and Total 315 

211. Remarks 319 

212. Geometric Representation of Partial and Total Derivatives 319 

CHAPTER XXIX 

MULTIPLE INTEGRALS 

213. The Volume under a Surface 321 

214. The Double Integral 322 

215. A Plane Area Expressed as a Double Integral 327 * 

216. The Area of a Surface Expressed as a Double Integral 328 

217. Exercises 331 

218. The Triple Integral 332 

219. The Triple Integral in Polar Coordinates 335 

220. Volumes by Triple Integration 337 

CHAPTER XXX 
CENTERS OF MASS 

221. Center of Mass of a Solid 338 

222. Center of Mass of a Plane Area 341 

223. Center of Mass of an Arc of a Plane Curve 342 

224. Center of Mass of a Surface of Revolution 343 

225. Center of Mass of Any Curved Surface 344 

226. Exercises 346 



BOOK V 

THEOREMS OF TAYLOR AND MACLAURIN. INTEGRA- 
TION OF RATIONAL FRACTIONS. ENVELOPES 

CHAPTER XXXI 

THEOREMS OF TAYLOR AND MACLAURIN. DE MOIVRE'S 
THEOREM AND THE HYPERBOLIC FUNCTIONS 

227. Expression of a Function as a Polynomial 349 

228. Exercises 355 

229. Taylor's Theorem: Finite Form 355 

230. Examples of Development by Taylor's Theorem 358 

231. Exercises 360 



XVI CONTENTS 

Article Page 

232. Maclaurin's Theorem: Finite Form . 360 

233. Exercises 361 

234. A Property of Polynomials 361 

235. Evaluation of the Indeterminate Form - by Taylor's Theorem .... 362 

236. Criteria for Maxima and Minima Determined by Taylor's Theorem 363 

237. Theorems of Taylor and Maclaurin: Infinite Forms 366 

238. Examples of the Application of the Infinite Forms of the Theorems 

of Taylor and Maclaurin 369 

239. Exercises 374 

240. The Power Series 374 

241. The Ratio Test for the Convergence of Power Series 377 

242. Exercises 378 

243. Development of/ (x) when a Formula for / (x) Cannot Easily be 

Found 378 

244. Exercises 379 

245. Computation of Logarithms 380 

246. Exercises 382 

247. Theorems of Taylor and Maclaurin for Functions of Any Number 

of Variables 382 

248. Exercises 386 

249. Euler's Theorem for Homogeneous Functions ■■ 386 

250. Exercises 388 

251. The Imaginary Exponent. De Moivre's Theorem 388 

252. Exercises , 392 

253. The Hyperbolic Functions 392 

254. The Inverse Hyperbolic Functions 393 

255. Geometric Properties of the Hyperbolic Functions 395 



CHAPTER XXXII 
INTEGRATION OF RATIONAL FRACTIONS 

256. Properties of Polynomials 398 

257. Integration of a Rational Fraction when the Denominator is Re- 

solved into its Linear Factors 399 

258. Exercises 402 

259. Case 2 402 

260. Exercises 403 

261. Integration of a Rational Fraction when the Denominator Con- 

tains Imaginary Linear Factors 403 

262. Exercises 405 

263. Case 4 406 



CONTENTS XV11 

Article Page 

264. Exercises 408 

265. The Integral of Any Rational Fraction 408 

266. Miscellaneous Exercises 409 

CHAPTER XXXIII 
ENVELOPES 

267. Systems of Curves 410 

268. Exercises 410 

269. Envelopes 411 

270. An Important Property of the Envelope 413 

271 . Exercises 415 

272. Caustics : 416 

273. Exercises 418 

BOOK VI 

AN INTRODUCTION TO ORDINARY DIFFERENTIAL 
EQUATIONS 

CHAPTER XXXIV 
ORDINARY DIFFERENTIAL EQUATIONS OF THE FIRST ORDER 

274. Definitions 421 

275. Solution. Particular Integrals. The Complete Primitive 422 

276. Derivation of the Differential Equation from the Complete Primi- 

tive 424 

277. Exercises 425 

278. Geometrical and Physical Interpretation 425 

279. Solving a Differential Equation 426 

280. The Differential Equation of the First Order and Degree 427 

281. Variables Separable 427 

282. Exercises. ... 428 

283. Variables Separable by Transformation 428 

284. Exercises 429 

285. The Integrating Factor 429 

286. Exercises , 431 

287. The Homogeneous Equation 431 

288. Exercises 433 

289. The Linear Equation 433 

290. Exercises 434 

291. Equations Reducible to Linear Form 435 

292. Exercises 436 

293. Clairaut's Equation 436 



J ,v 



XVlli CONTENTS 

CHAPTER XXXV 

THE LINEAR DIFFERENTIAL EQUATION OF THE 22th ORDER 

WITH CONSTANT COEFFICIENTS 

Abticle Page 

294. Definitions 438 

295. The Solution of (A) when the Second Member is Zero 438 

296. Exercises 440 

297. Complex Roots 440 

298. Exercises 441 

299. Multiple Roots 441 

300. Exercises 444 

301. The Solution of (A) when the Second Member is not Zero 444 

302. Exercises . . ' 447 

303. Two Special Equations 448 

304. Exercises 449 

305. Miscellaneous Exercises 449 

Answers 451 

Index 477 



THE 


GREEK ALPHABET 




A a al'pha 


I 1 io'ta 


P P 


rho 


B p be'ta 


K a kap'pa 


2 a 


sig'ma 


r y gam'ma 


A X lamb'da 


T r 


tau 


A 5 del'ta 


M/x mu 


T v 


up'silon 


E e ep'silon 


N v nu 


$ 4> 


phi 


Z f ze'ta 


S I xi 


x x 


chi 


Hq e'ta 


om'icron 


<fr ^ 


psi 


6 0$ the'ta 


II ir pi 


03 


omeg'a 



DIFFERENTIAL AND INTEGRAL CALCULUS 

BOOK I 
DIFFERENTIAL CALCULUS 



CHAPTER I 
INTRODUCTION 

1. Variables and Constants. A variable is a quantity which 
is free to take an infinity of values in the course of the investiga- 
tion. Familiar examples of variables are the coordinates x and 
y in the equation of a curve. 

A constant is a quantity which keeps the same value throughout 
the investigation. If its value is in the nature of things absolutely 
unchangeable, such as 2, log 4, t, the constant is termed an 
absolute constant. If the constant remains fixed but undetermined 
in value during the investigation, while free to take different 
values in different problems, it is termed an arbitrary constant 
Thus a, b, c in the equation of the right line ax + by + c = 0; 
a, b, r in the equation of the circle (x — a) 2 -f- (y — b) 2 = r 2 , 
are arbitrary constants. Arbitrary constants are usually repre- 
sented by letters in the first part of the alphabet, as a, 6, c, ... , 
and variables by letters in the latter part, as s, t, u, v, w, x, y, z, 
or by Greek letters. Also, letters affected with accents or sub- 
scripts are used to represent both constants and variables, such 
as a', a", a'", a IV , . . . (read "a prime," "a second," . . . ); and 
Go, (hj a 2 , . . ., x , x h X2, . . . (read "a zero," "a one," "a two," 
. . .). 

1 



2 DIFFERENTIAL CALCULUS §2 

2. Functions. When two variables, x and y, are so related 
that the value of y depends upon the value of x, y is said to be a 
function of x, and x is termed the argument of y. 

The function is also called the dependent variable, and the argu- 
ment the independent variable. 

For example, the volume of a sphere is a function of its radius, 
because it depends for its value upon the value of the radius. 
Also, the ordinate of a point on a curve is a function of its abscissa, 
the sine of an angle is a function of the angle, the velocity of a 
falling body is a function of the time that it has been falling, etc. , 
etc. 

On the other hand, since the radius of a sphere depends for its 
value upon the volume, the radius may be regarded as the func- 
tion and the volume as the argument. So, too, the abscissa of a 
point on a curve may be regarded as a function of the ordinate, an 
angle as a function of its sine, etc., etc. And, in general, if y is a 
function of x, x is also a function of y. 

A variable may be a function of several arguments. Thus, 
since the area of a rectangle depends in value upon its length 
and breadth, the area is a function of both length and breadth. 
Also, the slope of a tangent to a curve is a function of both coor- 
dinates of the point of contact; the volume of a rectangular paral- 
lelopiped is a function of its three dimensions; the velocity of a 
projectile is a function of the weight of the powder used in its dis- 
charge, of the time elapsed since the projectile was fired, of the 
resistance of the air, of the angle of elevation of the gun, etc., etc. 

If x represents a variable, every mathematical expression which 
contains x represents a function of x. Thus, 

(a) 3 x 2 + 2 x — 1, 4 ) tan re, sin -1 



re 4 + 3' 1+x,' 

2*, log(a;+ Vz 2 + l), 

all represent functions of x. 

Again, every equation between two variables expresses a func- 
tional relation between them. Thus each of the equations 



§2 INTRODUCTION 3 

(6) y = x sin x, y = 3 x 3 + x log x, 

b 2 x 2 + a 2 y 2 = a 2 b 2 , x cos y + y cos z + xy = 0, 
asserts that y is a function of x and that # is a function of y; for 
each equation asserts that either variable in it depends in value 
upon the other variable. 

When the function is explicitly given in terms of its argument, 
as is y in the first two equations of (b), it is called an explicit 
function. When the functional relation between the variables is 
implied, as in the last two equations of (6), the function is called 
an implicit function. 

The functional relation between two variables may be expressed 
in terms of a third variable termed a parameter. Thus, each of 
the following pairs of equations expresses a functional relation 
between x and y: 

x = t 2 — 1, x = a cos 6, x — a cos 3 0, 

y = t 3 + 2, y = b sin 6, y = b sin 3 0. 

For, y depends upon the parameter t or 6, which in turn depends 
upon x, and so y depends upon x. Similarly x depends upon y. 
The reader will perhaps recognize the second pair of equations as 
the parametric equations of the ellipse. They express the same func- 
tional relation between x and y that the equation b 2 x 2 + a 2 y 2 = a 2 b 2 
expresses. 

A function depends for its value not only upon its variable 
argument, but also upon constants. Thus, the volume of a sphere 
is not only a function of its variable radius, but also of the absolute 
constant, it; the function log a z depends for its value not only upon 
x, but also upon the base a; the equation b 2 x 2 + a 2 y 2 = a 2 b 2 
expresses that y depends in value not only upon x, but also upon 
a, b, and the exponent 2, and is a function of a, b, and 2. Every 
quantity is a function of all the quantities that are necessary to 
determine its value, but when we use the term "function" we usu- 
ally have in mind only the dependence of the function upon its 
variable argument. 

A function may be of such a nature as to take two or more 
different values for each value of its argument. For example, 



4 DIFFERENTIAL CALCULUS §3 

vV + 2 has two values for each value of x. Such a function is 
termed two-valued. There are three-valued functions, four-valued 
functions, etc., etc. Since there are an infinite number of arcs 
whose sine is a given number, sin -1 x is an infinitely many-valued 
function. So, too, are cos -1 x, tan -1 x, etc. • The several values of 
a many-valued function are sometimes termed its branches (the 
significance of the term will appear later), and may be regarded 
simply as so many distinct though closely related functions. 
Thus, + Vx 2 + 2 and — Vx 2 + 2 may be regarded either as the 
two values or branches of a two-valued function, or as two distinct 
functions which differ only in sign. 

A function which has but one value for each value of its argu- 
ment is called one-valued or single-valued. 

3. Functional Symbols. The symbol f(x) is used to denote 
any function of x whatever,* and is read "function x," or "func- 
tion of x," or " / function x." When several functions are under 
consideration, we distinguish them from one another by using 
different letters for the functional symbol. For example, 

fix), F(x), g(x), f(x), /"(a), /iW, /.(*), 0(x), *(*), 

represent different functions of x and are read "/ function x," 
"cap/ function x" "g function x," etc. That u is a function of 
x may be expressed by an equation such as 

u = f(x), u = fi(x), u = g(x). 

A function of two arguments x and y is denoted by such symbols 
as f(x,y), F(x,y), read "/ function of x and y," "cap / func- 
tion of x and y." That u is a function of x and y may be ex- 
pressed by an equation such as 

u=f(x,y) or u = <f>(x,y). 

* f(x) does not mean the product of / and x any more than tanf x + 1) 

means the product of t, a, and n by the binomial f x + -)• The / in the 

symbol f(x) stands for the word function, just as tan stands for the word 
tangent. 



§4 INTRODUCTION 5 

A function of three arguments x, y, z is denoted by f(x,y,z), 
F(x,y,z), etc. 

The equation f(x, y) = asserts a functional relation between 
x and y without specifying which is argument and which is func- 
tion. Similarly, such an equation as f(x,y,z) = asserts that 
z is a function of x and y, or that z is a function of y and z, or that 
y is a function of x and z. 

4. The Identity. An equation is an identical equation, or 
simply an identity, if, when all the indicated arithmetical opera- 
tions are performed, the two members of the resulting equation 
are alike in every respect. The sign of identity is = , which is 
read " is identical with, " " is identically equal to. " For example, 

(a) 3 - (2 - 1) . 7 - (3 + 2), 

x 2 - 3 x + 2 = (x - 1) (x - 2), 
4:X 2 -9y 2 =(2x+Sy)(2x-Zy), 

(x+ a)(x - a) - x2 + ^ 2 = 0, 

are identities. 

The characteristic property of an identity follows directly from 
the definition, and is this : 

An identity is true for all values which may be assigned to the 
unknowns or variables in it* 

This is seen to be true of identities (a). Herein an identity is 
distinguished from an equation which holds true for only a limited 
number of values of the unknowns or variables in it. For example, 

(6) x 2 - 3 x + 2 = 0, 4 x 2 - 9 y 2 = 5, 

are equations but not identities, and are not true for all values 
of x and y. By solving these equations we find that the first 
holds true only w hen x = 1 or 2, and the second only when 
x = ± % V5 + 9 y 2 , that is, only for certain pairs of values (infinite 
in number indeed) of x and y. 

* There are exceptions to this statement; there are identities which fail 
for certain values of the unknowns in them. 



6 DIFFERENTIAL CALCULUS §5 

If we have given such an identity as 

ax? + bx 2 + ex + d = Ax* + Bx 2 + Cx + D, 

that is, if we know in advance that this is an identity in x, then 

a = A, b = B, c = C, d = D. 

Again, if we know that 

ax? + bx 2 + ex + d = 0, 

then we can assert that 

a = 0, b = 0, c = 0, d = 0. 

All this follows from the definition of the identity. 

The sign of identity is often written between a symbol and the 
mathematical expression for which it stands, as in 

/ (x) = ax 2 + bx + c and cos x = Vl — sin 2 x. 

In such cases the sign = may be read as above, or it may be read 
"stands for," "is defined to be," and the like. 

Identities are frequently written as simple equations. Thus, 
identities (a) are usually written with the sign of equality =. 
The sign = is used whenever it is desired to call particular atten- 
tion to the absolute identity of the two expressions between which 
it stands. 

The following example and exercises are intended to familiarize 
the student with the use of functional symbols. 

Example. f(x) = x 2 - 3: find /(2), /(0), /(V§), f(-\ and/(o+ 1). 

w 

Solution. /(2) =2 2 -3 = 1, /(0) =0-3 = -3, /(V§) =(V3>-3 = 0, 

f(~) = h- S = LjZ ^'/(« + l) = (« + l) 2 -3 = a 2 +2a-2. 
\z ) z* z* 



5. Exercises. 

1. f(x) = x 2 -2x+2: find/CD, /(-l), /(2), /(J), /(0),/(a-l). 

2. /(*) = ^+x- 6: find/a), /(2), /&+ 2), /(6- 3), /^). 

3 - ( * ) = |rr! : find ? (2) > ^ (3) ' ^ ( - 3) ' ^ (1) ' *(«} 



§ 6 INTRODUCTION 7 

4. 0(?/) = -^±=L: find 0(0), 0(1), 0(-l), <t>(Vs), 0(tan0), and 
show that <t>[-) = <i>{x). 



♦@ 



5 - AW = Vrr : find /i(0) > /i(1) ' /i(2) > and show that 

3+1 



/i(£)--tfiM, and /^-^ 



+ /1W--I. 



— : find^f-V P(sin0), P(cos0). 

^2 VW 



6. F(t*)~- ^ 

vl 

7. P(<*) = — ^ : show that F (-) = F(a), and P(tan 0) = isin 2 0. 

1 + a 2 \a / 2 

8. P(z) = :r 3 -3z 2 +3x+5: findP(-l), P(0), P(l), P(x+1). 

9. /(jc) is of such a nature that /(a#) =f(x) +f(y). 

(a) Put ?/ = 1, and show that /(l)sO. 

(6) Put y = -, and show that /(-) = — f(x). 
x \xJ 

(c) Put y = x, and show that f(x 2 ) = 2f(x). 

(d) Show that f(x m ) = mf(x) when m is a positive integer. 
What simple function has the above properties? 

10. (x) is of such a form that 4>{x-\-y)= 4>(x) (y). 
(a) Put y = 0, and thus show that 0(0) = 1. 

(6) Put ?/= —x, and thus show that 0(— z) = — — • 

<t>{x) 

(c) Put 2/ = aj, and thus show that 0(2 x) = [0 (a:)] 2 . 

(d) Show that (ma;) = [0 {x)] m for all integral values of m, posi- 

tive and negative. 
What function has these properties? 

6. Kinds of Functions. (1) A polynomial in x is an algebraic 
sum of multiples of positive integral powers of x. The following 
are polynomials, 

2 z 5 - 6 x 4 + x 3 - 7; x 2 - 1; 3 z 4 + x 2 ; f z 3 - V^x 2 + ttx. 

The degree of a polynomial in x is the highest exponent of x in it. 
The foregoing polynomials are of the fifth, second, fourth, and 
third degrees respectively. A constant may be regarded as a 
polynomial of the 0th degree because k = kx°. 

The general type of a polynomial of the nth degree is 
ax n + bx n ~ l + cx n ~ 2 + . . . + hx 2 + kx + I, 



8 DIFFERENTIAL CALCULUS §6 

where a, b, c . . . h, k, I are constants and n is a positive integer. 
It is understood that n is finite, that is, that the polynomial has 
a finite (limited) number of terms. 

(2) A rational fraction in x is the ratio of two polynomials in x. 
For example, 

1 3^ x - 1 x 2 + x - 1 V7x*+w 

x x 2 ' x 2 + 3' x b -x 2 + 2' 3s 3 -2z + log3' 

are rational fractions in x. Polynomials and rational fractions 
are termed rational functions. 

(3) When a function which would otherwise be a polynomial 
or a rational fraction contains the argument under a radical sign 
or affected with a fractional exponent, the function is said to be 
irrational. For example, 

' v a; x 2 + xs — 2 
are irrational functions. 

A function in which the argument is affected with a negative 
exponent can be brought to the form of a fraction, rational or 
irrational, having positive exponents only. For example, 

x = ^> x + x *= —J ^ 2 + (*- i)" 3 = —^7 771 — 

x x 2 (x — l) 3 

The first of these is a rational fraction and the others are irrational 
fractions. 

Polynomials, rational fractions, and the irrational functions just 
described are the simplest cases of a very general class of functions 
known as algebraic functions. The complete definition of algebraic 
functions is beyond the scope of this book. 

All functions that are not algebraic are termed transcendental. 
In this book we shall meet with but four kinds of transcendental 
functions, and with these the reader is already more or less familiar. 
They are 

(4) Trigonometric functions, such as 

% 

sin x, cos x, tan x, sin (x 2 + 1), x sec V x 2 + 1, 3 x + tan , 

vz+1 



§7 



INTRODUCTION 



(5) Anti-trigonometric or circular functions, such as 



sin -1 a;, cos -1 



x, tan _1 z, sin -1 (:r + Vx 2 + l), - 



x + sec" 1 ^ 

Vx~^2 ' 



(6) Logarithmic functions, such as 

z 3 + log(2z 2 + x- 1) 



logo:, x -f log Vx + 1, 



log V: 



(7) Anti-logarithmic or exponential functions, such as 



5* 2+1 , (a 2 + iy 



(3 a- 2)- 



, z + z 2 (a -f- 1)*. 

7. Graphs of Functions. We may ascribe to the argument x 
of a function /(#) all values real and complex, and study the result- 
ing real and complex values of the function. Such a study of all 
values of a function for all values of its argument belongs properly 
to the general theory of functions, whereas we shall in this book 
confine ourselves almost wholly to the case where both argument 
and function are real. We are concerned, therefore, with real 
values of argument and function. 

Each pair of real values of x and f{x) may be represented in 
magnitude and sign by the coordi- 
nates of a point in the plane, the 
abscissa representing the value of x, 
and the ordinate that of f(x) . Then 
the successive pairs of real values 
of argument and function will be 
represented by the coordinates of a 
succession of points which consti- 
tute a curve. This curve is called 
the graph of the function, f(x), and 
its equation is y = f(x). Thus the 
curve in the figure is the graph of the function x 4 

equation is 

V = x 4 — 2 x 3 . 




2x 3 , and its 



The graph has the advantage of presenting to the eye the real 
values of function and argument and of thus enabling us to com- 



10 DIFFERENTIAL CALCULUS §8 

pare these values readily. And every property of the function has 
its representation in a property of the graph, so that the algebraic 
problem of determining the properties of the function is trans- 
formed into a geometric problem of determining the properties 
of a curve. 

Not only does every function have a graph, but every curve is 
the graph of some function. For example, the equilateral hyper- 

k k 

bola xy = k or y = - is the graph of the function - . The pa- 

X X 

rabola y = 4 x 2 is the graph of the function 4 x 2 . The parabola 
y 2 = 4 x is the graph of the two-valued function ± 2 Vx, the 
upper branch of the curve being the graph of + 2 Vx, and the 
lower branch that of — 2 Vx. So too the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 



is the graph of the two-valued function ± — v a 2 — x 2 , that part 



a 



+ - Va 2 — x 2 , and that part which lies below the z-axis being the 

CL 



of the curve which lies above the z-axis being the graph of 
+ - Va 2 — x 2 , ar 

CL 

graph of Va 2 — x 2 . 

a 

8. Discontinuities of Functions. Usually a function has a 
definite value (real or complex) for each value of its argument, 
but it may happen that, for certain values of the argument, the 
function has no value whatever, in other words, does not exist. 
If f(x) has no value when x = a, that is, if f(a) does not exist, 
f(x) is said to be discontinuous when x = a, or to have a discon- 
tinuity when x = a. Its graph will lack the point whose abscissa 
is a: there will be at that point a break in the continuity of the 
curve. The following are examples of functions with discontinu- 
ities. 

Bin T 

(1) The function has no existence when x = 0. 'For when 

x 

x = 0, the function takes the form - , which has no value because 
cannot be used as a divisor. The function has, therefore, a dis- 
continuity when x = 0. For all other finite values of x, has 



INTRODUCTION 



11 



a definite and determinate value. A clear insight into the nature 
of this singularity may be had by a study of the graph of the func- 
tion as shown in the figure. Its equation is of course y = 



sinz 



x 



For very small + and — values of x, 



sin x 

x 



differs little from 1, and 



the smaller the numerical value of x, the nearer does the function 




come to the value 1, so that the curve approaches as near as we 
please to the point (0, 1) from either side of the y-axis. The point 
(0, 1), however, does not itself belong to the graph: the graph 
may be regarded as broken in two at this point or as having had a 
hole punched in it here. This point is a singular point, or a point 
of discontinuity of the graph, or of the function. 

(2) The function — has no existence when x = 0. For when 



x = 0, the function has the form - , which has no value because it 

is impossible to divide 1 by 0. 
The function has, therefore, a 
discontinuity when x = 0, but 
has a determinate value for all 
other value's of x. The graph 
is shown in the figure. It has 
no point whose abscissa is 0, 
that is, it has no point in com- 
mon with the 2/-axis. When x 




is a very small + or 



number, — is a very large + number, and 

X" 1 



12 



DIFFERENTIAL CALCULUS 



§8 



the smaller x becomes the larger is -5 . We maj r say, then, that 

the curve is broken in two at the point whose abscissa is 0, but 
that, unlike the preceding curve, the break has receded along the 
2/-axis to an indefinitely great distance from the origin. We may 
here introduce the point at infinity or the infinite point of a line, 
which is a fictitious or ideal point whose distance from any finite 
point of the line is greater than any magnitude that can be named. 
We may then say that the infinite point of the ?/-axis is a point of 



discontinuity of the graph of -= or of the function 



1_ 

x 1 



(3) The function - is discontinuous when x 
x 



1 



because ^ has no 

value. The graph of this function is an equilateral hyperbola 

having the axes of coordinates 
for asymptotes as shown in the 
figure. The graph has no point 
whose abscissa is 0, but when x 



— X 



is a very small =L number, - is 

x 

a very large ± number. The 
graph is not only broken, as in 
the preceding examples, but, un- 
like them, the severed ends are 
widely sundered, so that the discontinuity is very evident. 



(4) Tan x has, in strict logic, no value when x = ± ~ , ± -~- 



etc. 



For, 



tan x is defined to be the ratio t of the legs of a 



right triangle. If we make x a right angle, the sides h 
and a are parallel and do not meet, and consequently 

neither a nor the ratio T has a determinate value. 



o 

Tan x has, therefore, discontinuities at x = ± ~, ± -~-, . 




x is a trifle 



I less 
( greater 



When 



than ± 



, tan x is a very large 



INTRODUCTION 



13 



j _ | number, so that if we draw the graph of the function it will 

be seen to consist of an unlimited number of separate branches. 
On this graph are no points whose abscissas are 



7T ,3 7T ,57T 

2' ~2~' ~2~' ' 



The graph is broken at each of these points, and, as in the graph 
of -, the severed ends are widely sundered. 

X 




The student must not infer from the foregoing discussion that 
the textbooks in trigonometry are in error in writing 



tanMz^j = oo,tan(±-^n =00, 



etc. 



It may well be that the student has himself fallen into the error 
of giving a too literal interpretation to these equations. For, 

7T 7T 

tan- =00 means, not that tan^ actually exists and has a definite 
value designated by the character 00 , but merely that tan x is 

IT 

indefinitely large when x is indefinitely near to the value — • We 
shall return to this point again in Art. 17. 

/v>2 — /^2 



(5) The function 



x — a 



has no value when x = a. For, in that 



sin x 

case, this function, like , takes the form - , which has no value. 

x 

The function has a discontinuity when x = a. Its graph may be 
determined in the following way. 



14 



DIFFERENTIAL CALCULUS 



§8 



We have the identity (Art. 4) 



= x + a, 



which holds true for all values of x except for x = a (see page 5, 
footnote). The reason that we can assert that this identity 

fails when x = a is that in 
that case the first member 
of the identity becomes non- 
existent, while the second 
member takes the value 2 a. 




m X Hence the graph of 



*2 _ 



IS 

x — a 

identical with the graph of 

x + a with the single exception that the former lacks the point 

(a, 2 a). The graph of x + a is the right line shown in the figure, 

°" 2 — n ^ 

is this same right line with the point 



and the graph of 
(a, 2 a) lacking. 



x — a 



(6) A part of the graph of the function 2 X is shown in the figure. 

There is a discontinuity at the origin because the function takes 

i 
there the form 2 ° , which has 
no meaning. The equation of 

this curve is of course 
i 
y = &. 

Now when x is + and very 

i 
small, 2 X or y is + and very 
large, and the smaller x becomes, 
the larger does y become. On the other hand, when x is — and 

i 
very small, 2 X or y is + and also very small, and the smaller x 
becomes, the smaller does y become. The left-hand branch of the 
curve terminates abruptly at the origin. It is plain that the curve 
is broken at x = 0, and that the severed ends are widely sundered, 




§8 INTRODUCTION 15 

one being at the origin and the other at the " infinite point" of 
the ?/-axis. 

As an exercise the student should form a table of values of x 
and y for this curve, using a table of logarithms, and should then 
draw the curve on coordinate paper. 



CHAPTER II 

LIMITS 

9. Definition of Limit. When a variable, x, approaches in 
value a constant, a, in such a way that the numerical value of their 
difference, x — a, eventually becomes less and remains less than any 
magnitude that can be named however small, a is defined to be the 
limit of x, and x is said to approach a as a limit. 

The following examples will aid in the understanding of this 
definition. 

Example 1. Suppose a point B to start from a position C in the 
z-axis and to move along that axis toward a fixed point A. To fix ideas, 
in the figure A is placed at the right of the origin with the abscissa a, and 
C is placed at a distance d to the right of A, so that the abscissa of C is 
a + d. Let x be the variable abscissa of B. We now impose upon B a 
law of motion, or, what is the same thing, we impose upon x a law of 
variation that shall cause B to move, in the first second of time, to d, the 
O A C Co c C ^^-point °f AC, in the next sec- 

J J — L ^— 1 - = ' *< <q**B ~atd on d to C 2 , the mid-point of AC h 

in the third second to C 3 , the 
mid-point of Ad, and so on. That is, in each successive second, B is 
made to traverse half the distance that separates it from A. We assert 
that x has the limit a as defined above. For, at the end of successive 
seconds, the difference x — a has the values 

*J X A X A l J 1 r1 

2 d > 2* ' 2* ' & ' ' ' ' ¥ 

Now let e be any small positive quantity whatever. Plainly, we can 
always find a term in the above series which shall be less than e, and all 
succeeding terms will be less than e.* 



* For example, let e = * -d. Then, since 2 20 = 1,048,576, -^d <«. 

That is, at the end of the twentieth second x — a is less than this e, and will 
remain less for all ensuing time. 

16 



§ 9 LIMITS 17 

Therefore, a is the limit of x. 

The fixed point A is termed the limiting position or the limit of B, and B 
is said to approach A as its limiting position or as its limit. 

Example 2. In the foregoing exaniple we have a variable which is 

always greater than its limit and decreases towards that limit. In the 

following figure we have placed C at a distance d to the left of A. Then, 

if B, starting at C, move toward 

A as before, traversing in each -* ■ = — — ^ ,2 |3 ' 

a-d B" *" a 

second just half the distance that 

separates it from A, B will have A as its limiting position, and x will have 
a as its limit. For, at the end of successive seconds, the difference a - x 
will have the same series of values that x — a has in the preceding ex- 
ample, and the same argument applies as in that case. Here we have a 
variable which remains less than its limit and increases towards that 
limit. 

Example 3. There are many other laws of motion which would cause 
B to approach A as its limiting position, and x to approach a as its limit. 
Thus, let B start at C, a point at a distance d to the right or left of A as 
before (see the figures of the preceding examples), and move so that its 
distances from C at the end of successive seconds are 

1 , 2, 3,4, n j 

2 d > 3 d > i d > 5 d > ' ■ ' ^Tl d " ' ' 

Then the series of values of x — a (or a — x) is 

1 , 1 , 1 , 1 , 1 , 

2 d > 3 d ' i d > 5 d > * ' ' ~n d > ' ' ' 

and it is plain that, however small e may be, we can always find a term in 
the last series such that this and all succeeding terms shall be less than e. 
Then a is the limit of x, and A the limiting position of B. 

Problem. Suppose that B covers one gth of the distance that separates 
it from A in each successive second. Show that B has A as a limit, no 

matter how small a fraction - may be. 

Q 

In each of the forgoing examples x — a (or a — x) never becomes abso- 
lutely zero, and x is never quite equal to its limit a. B never traverses 
the entire distance that separates it from A, and hence never comes into 
actual coincidence with A. In the following example B comes into coin- 
cidence with A, and x becomes equal to a. 

Example 4. Let A and B be upon the z-axis as before, and let M and 
N be fixed points on either side of A. Let B start at M and move in the 
first second to Ni, the mid-point of NA, in the next second in the reverse 



18 DIFFERENTIAL CALCULUS §10 

direction to Mi, the mid-point of MA, in the third second back to N 2 , the 
mid-point of NiA, then to M 2 , the mid-point of MiA, and so on. B swings 

O N Ni N 2 A M 2 Mi M backwards and forwards through 

A, and x becomes equal to a once 
at each oscillation. In accordance with our definition, a is the limit of x, 
and A is the limiting position of B. In this case the variable is alter- 
nately greater and less than its limit. 

It should be noted in each of these four examples that a is not 
the final value of x and A is not the final position of B, because the 
variation of x and the consequent motion of B are arbitrarily 
determined in such a way that x can have no final value and B 
no final position. We may say, what is in fact implied in our 
definition of limit, that : 

The limit of a variable is not an ultimate value which the variable 
reaches, but a value which it strives to reach, and which it suggests. 

That a is the limit of x is expressed thus : 

lima; = a, 
which is read " limit x equals a," or thus: 

x = a, 

which is read " x approaches a (as a limit)," the words in paren- 
theses, being always understood when not expressed. 

10. Infinitesimals and Infinites. Definition. — An infinitesimal 
is a variable whose limit is zero. 

An equivalent definition is the following: 

An infinitesimal is a variable which eventually becomes less numer- 
ically and remains less than any quantity that can be named however 
small. 

If, in the examples of Art. 9, a is 0, £ is an infinitesimal. In 
that case B has the origin as its limiting position. When a is not 
0, it is obvious that x — a (or a — x) is an infinitesimal. It follows 
from our definition of limit and of infinitesimal that: 

The difference between a variable and its limit is an infinitesimal. 

The word infinitesimal is used here in a technical sense. An 
infinitesimal as here defined is not necessarily a very small quantity. 
At some stage of its variation it may be very large. Its charac- 



§ 10 LIMITS 19 

teristic property is that it must eventually become, and remain, 
very small. 

After what has been said concerning limits, it is scarcely neces- 
sary to remark that a variable does not necessarily have a limit, 
and that it will not have one unless its law of variation be suitably 
chosen. For example, suppose B to start at M (see figure of Ex. 4, 
Art. 9), and to move first to N, then back to M , then to N again, 
and so on, continually oscillating between M and N. Plainly, x 
will have no limit. Again, suppose that B starts at the origin and 
moves in the positive direction along OX with uniform velocity 
(i.e., covering equal spaces in equal intervals of time). Obviously, 
x has no limit, but becomes in time greater than any magnitude 
that can be named. In this case x is an infinite. 

Definition. — An infinite is' a variable which eventually becomes 
greater (numerically) and remains greater than any magnitude that 
can be named however great. 

An infinite as here defined may at some stage of its variation 
be a very small quantity. Its characteristic property is that it 
must eventually become, and remain, very large. 

In spite of the fact that an infinite is a variable that has no limit 
whatever (the word infinite means without limit), usage sanctions 
a certain inaccuracy of language in this connection. We intro- 
duce the fictitious or ideal number, "infinity," which we treat in a 
measure as though it were a definite and determinate number, 
and represent by the symbol oo . We then say that an infinite is 
a variable whose limit is " infinity," and express this by the sym- 
bols, 

lim x = oo , or x = oo . 

As for the moving point B whose abscissa is the infinite x, we here 
introduce again (Art. 8, Ex. 2) the fictitious or "ideal" point of the 
£-axis, the "point at infinity," and say that the point at infinity 
on the z-axis is the limiting position of B. 

Problem. 

Show that the reciprocal of anjj^f^j is anjj^^,]. 



20 DIFFERENTIAL CALCULUS §11 

11. Limits of Functions. It has already been remarked that 
whether a variable has a limit or not depends upon its law of vari- 
ation. The problem that oftenest presents itself is this: We are 
given a function, f(x), and we make x approach some arbitrarily 
chosen limit, a. The law of variation of f(x) is determined by 
that of x, and the problem is to ascertain whether / (x) has a limit, 
and, if so, what that limit is. If, when x has the limit a, f(x) has 
the limit A, this fact is expressed thus: 

lim f(x) = A, 

x=a v 

which is read, "As x approaches a, the limit of f(x) is (or equals) A," 
or thus: "As x approaches a, limit /(x) is A." 

Example 1. Let f(x) = 5 x 2 and let x be made to approach 3 as a limit. 
To determine lim f(x) . 

Solution. We see intuitively that 5 x 2 has the limit 5.3 2 = 45. To 
show that this is in accord with our definition we proceed as follows : 

Let x = 3 + M then 5 x 2 = 5 (3 + k) 2 = 45 + 5 (6 + k) k. Asx = 3,k = 0, 
and by taking a value of x sufficiently near its limit 3, that is, by taking 
k small enough, we can make 5 (6 + k) k, which is the difference between 
5 x 2 and 45, less than any small number e, and this difference will remain 
less than e throughout the subsequent approach of x to its limit 3. Hence 
45 is the limit of 5 x 2 , and we may write 

lim 5 x 2 = 45. 

2 = 3 

Example 2. Determine the limits of sin x and cos x as x = 0. 

Solution. To fix ideas, we take x at such a stage in its variation that 
the angle x is acute. Then from the accompanying 
figure it is geometrically evident that as x = (with- 
out ever becoming quite equal to 0) the difference 
between sin x and on the one hand and cos x and 
1 on the other will eventually become less and will 
remain less than any namable magnitude however 
small. That is, 

lim sin x = and lim cos x — 1. 

Z=0 2=0 

The following are more difficult examples of the process of find- 
ing the limit of a function. The results obtained in examples 3 
and 4 will be of use in future chapters. 




§11 



LIMITS 



21 



Example 3. To determine lim • 

0=0 d 

Solution. If in this function we write for 9, the function takes the 

meaningless form - . Hence we cannot here determine the limit by simply 

substituting the limit of the argument in the place of the argument, as 
could have been done in the preceding 
examples. In the unit circle consider 
an arc AC = 9, which for convenience 

we suppose to lie between and |- 

AD, BC, and OB are respectively tan 9, 
sin 9, and cos 9. From the figure it is 
obvious that 

area OAD > area OAC > area OBC. 
By geometry 



Hence 




area 


OAD 


= \ tan 0; 


area 


OAC- 


= i 0; 


area 


OBC 


= \ sin 9 cos 9. 


tan 9 > 9 


> sin 9 cos 9, 



> - — > COS 9, 

cos 9 sin 9 



n ^ smfl 
cos 9 < < 



cos 9 
1 



Now let 6 = (C = A): then cos 9 = 1, and = 1. 

' cos 9 



Hence, since 



sin 9 



lies always between cos 9 and , the limit of the function is also 1. 

cos 9 

That is 

I» lim —3— = 1. 

9=0 e 



Example 4. To determine lim 

0=0 



1 — cos 9 



Solution. There is no loss of generality in taking 9 at such a stage of 
3 variation that either < 
cos 9 is positive. We have 



its variation that either < 9 <~ orO>0>-^,so that in either case 



1— cos 



1— cos 2 fl 

1 + COS 9 



sm 2 fl 
1 + cos 9 



< sin 2 9. 



22 DIFFERENTIAL CALCULUS §12 

Further, sin 6 < 6 and sin 2 6 < 6 sin 6. Hence, because 1 — cos d is positive, 
we have 

i cos 

< 1 — cos e < 6 sin 0, and < < sin 6. 

e 

But lim sin 6 = 0; therefore, 

0=0 

-r-r «• 1 — cos© - 
IT. lim s = 0. 

12. The Difference between the Limit of /(a?) when x = a, 
and the Value of /(«) when as = a. Formulae I and II of ex- 
amples 3 and 4 of the preceding article do not assert that — — = 1 



and = when = 0. Indeed, the question as to the 

o 

values of these functions does not arise here, because 6 in approach- 
ing as a limit is expressly excluded from taking the value 0. As 
a matter of fact, neither function has any value when 6 = 0. In 
being nonexistent that is, discontinuous, when 6 = 0, while having 
determinate limits when 6 = 0, these functions differ radically from 
those of examples 1 and 2 of the preceding article. For the functions 
5 x 2 , sin x, and cos x not only have the limits 45, 0, and 1 respec- 
tively when x = 3 and respectively, but they also take the values 
45, 0, and 1 when x = 3 and 0. It appears, then, that there are 
two cases to be considered. 

(1) lim f(x) = value f(x) = f(a). 

x=a x=a 

(2) lim f(x) 5* value f(x). 

x=a x=a 

In case lim f(x) is determinate while value f(x) is not, that is, in 

x=a x—a 

case (2), the limit is often used in place of the missing value of the 

sin x 
function. Thus, we may if we choose use lim , or 1, in place 

z=0 % 

sin x 
of the nonexistent value , and may thus ignore the discon- 

x = a X 

tinuity of this function. In such a case the function is said to have 
a removable discontinuity. Not every discontinuity is removable. 

Thus, in the case of the function of example 6, Art. 8, lim 2 X = 

*=o 



§ 13 LIMITS 23 

or oo , according as x approaches its limit, 0, from the left or from 
the right, that is, according as x remains always less than its limit, 
or always greater than its limit. In this case we cannot deter- 
mine whether to substitute or oo for the missing value of the 

function. This discontinuity of 2 X is not removable. 
In case (1), f(x) is continuous at x = a. 

13. Further Examples of Finding Limits of Functions. 

Example 1. To show that lim = -. 

0=0 & * 

Solution. From the identities 

1 — cos d 2 sin 2 



he 1 /sinj_0\ 
~2\ he I 

follows lim^J^ = ilim( s -^Y. 

0=0 2 2 0=0 V \e J 

But by example 3, Art. 11, lim S11 * 2 = 1. Therefore, 

0=0 i e 

r l — cos e l 

x 2 -a? 



Example 2. To determine lim 

x=a 2J — Oi 



Solution. The identity = x + a holds for all values of x save 

x— a 

for x = a. But x in approaching its limit is expressly excluded from 

taking the value a, and consequently this identity holds for all values 

of x that enter into the problem before us. We have, therefore, 

lim = lim (x+a) = 2a. 



The functions of examples 1 and 2 belong to the second class of func- 
tions described in Art. 12, that is, they are cases in which 

lim/(z) ^/(o). 

x=a 



24 DIFFERENTIAL CALCULUS §§14-15 

14. Exercises. Find the following limits: 



1. lim-^-. 

0=0 cos 

2. Iim-£2i£ 



=2 2 * 



£frnf. Put a; = - 
2 

3. lim-r-A- • 



x=osin -1 a; 

Hint. Put w = sin -1 x. 

-1 2~~* 



8. 


lim 1 -. 008 *. 
e=o sin 


9. 


t sin ax 
lim— 

2=0 ox 


10. 


,._ tan x — sin a; 


i=0 x 3 


11. 


lim***". 

z =o £ + a 


12. 


i. x 3 — a 3 
lim • 



z=a 



a:— a 



5- lim^- 13 . Em ^=4- 

1=0 * _ x = a x 2 -a 2 

a r 1 — sin a; 

x=o cos a: 14< H m *__«_. 

7. lim l-sina;. - z ~ a 

7T COSX 

* = 2 

i/mJ. Multiply both terms by 
1 + sin x. 

15. Meaning of the Symbols -~. = oc and — =0. In the prob- 
lem in Art. 10, it was seen that the reciprocal of an \ . _ . 1 

is an \ ! n „ n * e . A. These facts may be expressed thus 
( infinitesimal ) 

lim - = oo , and lim - == 0. 

And, a being a constant, we have the more general formulae, 

(a) lim - = oo , and lim - = 0. 

2 = % X = <X> % 

In place of these equations, usage permits us to employ for sake 
of brevity the symbolic equations, 

(6) - = oo , and — = 0, 

oo. ' 

expressions which have no meaning when interpreted literally, and 



uallyj; 



§ 16 LIMITS 25 

which are always to be understood as abbreviations for equations 
(a). Equations (a) or (6) are merely symbolical expressions of 
the obvious truth that if the denominator of a fraction contin- 

} numerically, while the numerator remains con- 
mcreases ) J ' 

stant, the fraction itself continually \ , > . 

( decreases ) 

16. The Limit of a Rational Fraction in oc, as a? = oo . Con- 

3 x 2 I ^ 

sider the fraction - — 5 ~ — , and let us determine its limit 

4 x 3 — 2 x 2 + 3 

when # = 00 . Dividing both numerator and denominator by x 3 , 

the highest power of x that occurs in the fraction, we get as an 

equivalent form 

M 

£ ar 



4-M 

X X 3 



Now when x = co , the numerator of this fraction has the limit 0, 
and the denominator the limit 4, by Art. 15. Hence 

,. 3a; 2 +1 n 

l 1 i24^-2^ + 3 = 4 = a 

2 a; 4 — 1 
Again, let us determine the limit of _ , r— = as a; = 00 . 

3 x 2 + x + 7 

Dividing both terms of the fraction by a; 4 we get the equivalent 
form 



. i.+i+l 

a: 2 a; 3 a; 4 

and we see that as x = 00 the numerator has the limit 2 and the 
denominator the limit 0. Hence the fraction has the limit 00, 
and we may write 

2rr 4 -l 2 

^3a, 2 + z + 7 = 6 =0 °' 



26 DIFFERENTIAL CALCULUS §17 

2 x 3 x -f- 3 

Finally, let us determine the limit of . _ as z = oo . 

o 2^ -j~ o x — 1 

Putting the fraction into the form 

2-1+3 

X 2 ^ x 3 

z a; 3 

we see that as x = oo the limit of the fraction is f . Hence 

2X 3 -£ + 3 _ 2 
S 5 x 3 + 3 x 2 + 1 ~ 5 

Let the student find the limit in the first of the foregoing exam- 
ples by dividing both terms of the fraction by the highest power 
of x that occurs in the numerator, and, in the second example, by 
dividing both terms by the highest power of x that occurs in the 
denominator. 

By a similar line of argument we can prove the general case and 
show that 

ax r J r bx r - 1 + . . . + hx + k 



lim 



a'x n + b'x n ~ 1 + . . . +h'x+h' 



is 0, oo , or — > according as r < n, r > n, or r = n. The student 

should give the proof in full. 

17. Infinite Limits of Trigonometric Functions. Tanz not 

TV 

only has no value when x = - (see Art. 8, example 4), but also 

7T 

no limit, properly speaking, when x =-. For in the latter case 

tan x increases (numerically) without limit, in other words, is an 
infinite. Here, too, we introduce the ideal number oo , and write 

7T 

lim tan x = oo , which is further abbreviated to tan - = oo . This 

.IT ^ 

Z = 2 

last expression is merely an abbreviation for the entirely accurate 
statement, "as x = - t tana; increases (numerically) without limit." 

q 

Similar explanations apply to such expressions as tan — - = oo , 
cot = oo , sec - = oo , and esc = oo . There is an ambiguity in 



§ 18 LIMITS 27 

these expressions, because, for example, lim tan x = + oo or — oo 



according as x = - from values less than ~ or greater than - • It is 
purely an arbitrary convention that we write tan ~ = + <» rather 
than — oo . Similar remarks apply in the other cases.* 

18. Example of a Function That Has No Limit. The function 

TV 

sin - has no limit when x = 0. v 
x 

TT 

For, in that case, - increases 

x 

7T 

without limit, while sin- oscil- o 
x 

lates between the extreme val- 
ues + 1 and — 1, passing through 
at each oscillation, but ap- 
proaching no limit. To con- 7 ' T '~ 
struct the graph of the function, we construct a table of values 




of x and sin 


x 


as follows: 










x = 4, 




4 

2' 


4 4 
3' 4' 


4 
5' 


4 
6' 


4 

7' 


4 

_, . . . 


7T 7T 

x ~ 4 




7T 

— > 

2 


3tt 

t TT 

4 *' 


5tt 
4 ' 


3tt 

T 


7?r , 
4 ' 


2x, . . 



sin- = jV2, 1, JV2, 0, -|V2, -1, -|V2, 0. . . . 

x 

As a: continues to approach 0, this series of values of sin- is re- 

x 

peated again and again, showing that the function has no limit. 

* In the text the inaccuracy of giving literal interpretations to such ex- 
pressions as ^. = oo, tan - = oo, is dwelt upon in order to bring out more 

clearly the nature of a limit, and to teach the student to think accurately, 
and not because there is much danger of inaccurate results arising from the 
literal interpretation of these expressions. Errors seldom arise in trigono- 
metric work from the literal interpretation of tan - = oo , sec 5 = 00, etc. 



28 . DIFFERENTIAL CALCULUS §§19-20 

In the figure only the right-hand branch is shown. A symmetrical 
branch lies at the left of the origin. 
Let the student show that: 

(1) The graph cuts OX in the points x = t where k is an integer. 

2 

(2) The graph touches the lines y = ± 1 at the points x = 7^ — -• 

(3) Therefore, between the y-axis and a line x = a are to be 
found an infinite number of arches of the graph, and that, too, 
no matter how small a may be. 

Problem. Show that tan- has no limit when x = 0, and draw the graph 
of the function. 

19. A Geometrical Limit. Let P and Q be two curve points. 
Let P remain fixed, while Q traverses the curve in such a way as 

to approach P as its limiting posi- 
tion. Then the secant PQ will turn 
about P as a pivot, and will approach 
a limiting position, TP, and this 
limiting position we define to be the 
tangent to the curve at P. 

The tangent is sometimes de- 
scribed (though inaccurately) as a 
line that joins two infinitely near points of the curve. It is plain 
that, as Q approaches P, the angle a which the secant makes with 
OX approaches, as a limit, 6, the angle which the tangent makes 
with OX. 

20. General Theorems of Limits. We shall now state some 
general theorems of limits, some of which we have made tacit 
use of in the foregoing pages. These theorems admit of rigorous 
proof, but these proofs do not belong to an elementary course in 
the calculus. In fact, most of these theorems are fairly obvious 
in the case of the simple functions that we shall deal with. 

a. The limit of f(x) is independent of the law of variation by 
which x approaches its limit. 




§20 LIMITS 29 

This means that whether x approaches its limit in such a way 
that the difference x — a (or a — x) has the series of values 
(Art. 9, example 1), 

2 d > 2> d > 2> d > * ' * 
or the series of values (Art. 9, example 3), 

1 j 1 a 1 a 

-d, -d, jd, . . . 

or some other series of values, f(x) has in each case the same limit, 
or if f(x) has no limit in one case it has none in any other case. 
There are exceptions to this principle, some of which are illustrated 
in the following examples. 

(1) Lim = +oo or — oo , according as x in approaching a is 

x—a X (Z 

always greater or less than a. 
i 

(2) Lim C x ~ a = oo or 0, according as x remains greater or less 

x=a 

than a. 

(3) Lim tan x = + oo or — oo , according as x remains less or 



7T 

greater than - (see Art. 17). 
A 

13. If two functions of the same argument are equal for all values 
which that argument takes in approaching a prescribed limit, their 
limits are equal, or if one function has no limit, the other has none. 

In symbolic language this principle takes the form: 

If f(x) = g{x), then lim /(as) = lim g(x). 

x=a x=a 

y. The limit of the sum or difference of two functions of the same 
argument is the sum or difference of their limits. 
In symbols, 

lim [f(x) db g(x)] = lim /(as) db lim g(x). 

x=a x=a x=a 

This theorem may be extended to cover the case of the algebraic 
sum of any finite number of functions of the same argument. 



30 DIFFERENTIAL CALCULUS §20 

5. The limit of the product of two functions of the same argument 
is the product of their limits, unless one limit is and the other oo . 
Expressed in mathematical symbols, this theorem becomes 

\imf(x) g(x) = lim /(re) • limgf(x). 

x=a x=a x—a 

This theorem can be extended to the product of several functions. 
In the case of the exception, the product of the limits and oo 
would have no value, whereas the limit of the product might be 
determinate. For example, 

(1) lim cos 6 tan 6 ^ lim cos • lim tan 0, 

„ . T „ . T „ . IT 

^2 ° = 2 6 =2 

because lim cos = 0, and lim tan = oo , and X oo has no value, 

6 ~2 6 ~2 

whereas cos 6 tan 6 = sin 6, and therefore 

lim cos 6 tan 6 = lim sin = 1. 

6 ~2 6 —2 

(2) lim — — t* lim sin 6 • lim - , 
0=0 c/ e=o 0=0 C7 

as is evident on inspection and by reference to Art. 11, example 3. 

As a corollary to this theorem we have the following, 
Corollary lim Cf(x) = C\imf(x), 

x=a x=a 

when C is a constant. 

€. The limit of the ratio of two functions of the same argument is 

the ratio of their limits, except when these are both or both oo . 

That is 

lim/(x) 

lim ;W - x=a 



= a 0(z) limg(x) 



x=a 



except when the limits of f(x) and g{x) are both or both oo . For 
example, 

lim sin 6 

lim sinj? e f\ = 1 = 2 

Q=\ 6 lim 6 ir ir 

-I 2 



§20 LIMITS 31 

In case the limits of j(x) and g{x) are both or both oo , their 

ratio would be ^ or — , neither of which has any meaning, whereas 

f( x ) 
the limit of the ratio ^^-i- might have a definite value. For 

example, 



sin e=o 



lim sin 



(1) lim— — F^ 

0=0 6 lim 

0=0 

because the second member has the meaningless form-, where?,s 
the first member is 1 (see Art. 11, example 3). 



X Cl x—a 



lim (x 2 — a 2 ) 



(2) lim 9* v 

x=a x — a lim (x — a) 

x=a 

for similar reasons. 



lim tan 



/o\ r tan 0*~ 2 

(3) lim 77^B^TT 



sec lim sec J 

because the second member has the form — , whereas = sin0, 

oo sec 

whose limit, as = ^, is 1. Nearly all the exercises of Art. 14 are 

cases of the exception to e. 
In case lim g(x) is or oo while lim f(x) is finite and different 

x=a x=a 

from 0, the theorem still holds by virtue of (6), Art. 15. 

The foregoing are the chief theorems of limits that we shall use 
in this course, although we shall have occasional need of the two 
following : 

r. Lim [f(x)] n = [lim/0r)l n , 

x=a Li=a J 

which is, in fact, a case of extension of 8. 
y). Lim log f(x) = log lim f(x). 



CHAPTER III 

THE DERIVATIVE 

21. Increments. Let Xi and xi + h be two values of x. The 
quantity h, which may be either + or — , is termed an increment of 
x. An increment of a variable is the amount of change it under- 
goes in passing from one value to another: it is the difference 
between two values of the variable. When x changes from X\ 
to xi + h, f(x) changes from f(xi) to f(x± + h), and the amount of 
this change which may be denoted by k is termed the increment of 
the function corresponding to the increment h of x. We have then 
the identity, or definition of k, 

k^ffa + h)-^) 
Instead of using h and k to denote the increments just defined, we 
shall usually employ the symbols Ax and Af(x), the first of which 
is read " increment x" or " 'delta x," and the second " increment 
function x " or " delta function x." Moreover, we shall usually 
omit the subscripts and write x, x + Arc for the two values of the 
argument. Then the foregoing identity takes the form 
(i) Af(x)^f(x + Ax)-f(x)* 

If y denote the functional symbol, the increment of y is Ay. Let 
x, f(x) be the coordinates of a point P on the graph of f(x) . The 
abscissa of a second point Q may be denoted by x + Ax, and then 
the ordinate of Q will be denoted by f(x + Ax) or f(x) + Af(x). 
The line segments representing Ax, Af(x), etc., are shown in the 
accompanying figures. In the second figure Af(x) is negative. 

* A result of this omission of the subscript is that we use x and f(x) sometimes 
to denote argument and function in general and sometimes to denote fixed 
values of them. This will occasion no difficulty to the careful student, and the 
use of the subscript would cumber the formulae unnecessarily. 

32 



22 



THE DERIVATIVE 
Y 



33 




f(x)+Wx) 



X O 




If y be the functional symbol, the coordinates of P are x and y, 
and those of Q are x + Az and 2/ + Ay, and the notation of the 
figures is as here shown. In the second figure Ay is negative. If 




a be the angle which the secant PQ makes with OX, it is seen 
from the figures that 



A/(s) = f(x + Ax)-f(x) 

Ax " Ax 



tan a, and -r-^ = tan a. 
Ax 



22. The Derivative. The fundamental idea of the Differen- 
tial Calculus is the derivative of a function, which is thus defined : 
The derivative of a function is the limit of the ratio 

increment of function 

— ^— — — — - — — - — — 9 

increment of argument 

as these increments approach the limit zero. 

The derivative of f(x) is denoted by f'(x) or Df(x), and the de- 
rivative of y by y' or Dy, the D standing for the word derivative. 



34 



DIFFERENTIAL CALCULUS 



§22 



Using the notation for increments just explained, we may express 
our definition as follows: 

/(») or I>m - lim *&*. - lim ** + *? ~ *°) , 

Aa;=0 AiC Ax=0 Aa? 

At/ 

y' or Dy = lim — • 
a*=o Ax 

We have just seen that 

A/0) Ay 
-ir^ or -~ = tan a. 
Ax Ax 

If 6 be the angle made with OX by the tangent PT, we know 

that as Q = P, the secant QP = the tangent PT, and a = 6 (see 

Art. 19). But when Ax = 0, Q = P. Hence 

.. A/O) ,. A?/ .-. , 

lim -4 — - or lim -r— = lim tan a = tan d, 

Az=0 AX Ax = oA£ Q=p 

or f'(x) or ?/ = tan 6. 

This important result may be expressed in words as follows: 

The geometrical interpretation of the derivative of /(as) 

Y 
Y 




= y-\-Ay 

O 



Af(x)=Ay 




f(x +Ax) 
= y+Ay 

is the slope of the tangent to the graph of f(pc) at the point 
(v,f(a>)). 

Example 1. To find the derivative of x 3 . 
Solution. By definition 



Dx 3 = lim 



Ax A 



From the identity 
we have 



Az=0A£ 



Af(x)=f(x+Ax)-f(x) 



AX 3 = (X + Az) ; 



= 3 x 2 Ax+ 3 x(Ax) 2 + (Ax) 3 *. 

* We use Ax n or A(z n ) to mean increment of x n , and (Ax) n to mean the 
nth power of Ax. Similarly, Dx n or D(x n ) means the derivative of x n , and 
(Dx) n means the nth power of Dx. 



Ax 3 ^ 
Ax 
AX 3 



lim — = lim (3 x 2 + 3 xAx + (Ax) 2 ) = 3 x 2 . 

Az=()AX Az=0 

Z)x 3 = 3 x 2 . 



§22 
Hence 

and 

Therefore 

It will add to the reader's 
understanding of this process to 
note the geometrical interpreta- 
tion of the successive steps. The 
graph of x 3 is shown in the figure. 
The coordinates of P are x and 
x 3 , and those of Q are x + Ax and 
(x + Ax) 3 . From the figure it is 
obvious that 

Ax 3 = (x + Ax) 3 — x 3 , 

A ., . Ax 3 QC , 
and that — = — = tan a. 
Ax PC 



THE DERIVATIVE 

= 3x 2 + 3xAx+(Ax) 2 , 



35 




(x+Ax)'' 



Hence 
and 



— = 3x 2 + 3xAx + (Ax) 2 

AX 

,. Ax 3 
lim — 

Ax=0 AX 



tan a, 



3 x 2 = lim tan a = tan 0, 



or 



D X 3 _ 3 X 2 = tan 0> 



Example 2. To find the derivative of 
Solution. By definition 









7)i = lim-^. 

£ Ax = AX 


From the 


identity 


A/(x)=/(x+Ax)-/(x) 


we have 




*i- 


112 xAx + (Ax) 2 


(x+Ax) 2 x 2 x 2 (x+Ax) 2 


Hence 






x 2 _ 2 x + Ax 

AX " X 2 (x+Ax) 2 ' 


and 


lim 

Ax=C 


Ax 


,. 2 x+Ax 2x 2 
ax=o x 2 (x 2 + Ax) 2 X 4 ~ X 3 


Therefore 






X 2 X 3 



by e, Art. 20. 



36 DIFFERENTIAL CALCULUS §23 

Example 3. To determine the derivative of 

Solution. Let f(x) = 

Then Af(x) ^ fix + Ax) - f(x) 



1 + x 2 



1 + x 2 

X + AX 



Hence 



l+(x+Ax) 2 1 + X 2 

(1 — x 2 )Ax— x(Ax) 2 

" (l+X 2 )[l+(x+Ax) 2 ]' 

Af(x) 1 — X 2 — XAX 

AX ~ (1 + X 2 ) [1 + (x + Ax) 2 ] ' 

, y Af(x) _ y 1 — X 2 — XAX 1 — X 2 

So AX ~ A% (l + X 2 )[l + (x+Ax) 2 ]~ (1+X 2 ) 2 

by e, Art. 20. 

Therefore finally D —f— = ~ ^ • 

1 + x 2 (1 + x 2 ) 2 

From the study of the foregoing examples, it appears that the 
process of finding the derivative of f(x) falls naturally into three 
steps : 

First step. The calculation of A f(x) by means of the identity 
A/0) ^ f(x + Ax) -fix) . . . (see (t), Art. 21). 

Second step. The formation of the function ^r — by dividing 
the result of the first step by Ax. 

Third step. The determination of lim — { — ■ =f f (x). In gen- 

Az=0 lAX 

iAfix) 
eral both x and Ax enter into the expression for -jr — , but through- 
out the process of finding the derivative x is kept constant, while 

Af(x) 
the variable argument of > is Ax. This step is often the most 

difficult of all, because the process of finding a limit is one which 
is apt to present serious difficulties. 

23. The Equations of Tangent and Normal. The derivative 
f'(x) or Df(x) is itself a function of x, and by giving different values 
to the x in it we get different numerical values of the derivative, 
that is, of the slope of the tangent at successive points of the graph. 
It is thus a simple matter to write down the equation of the tan- 



§24 THE DERIVATIVE 37 

gent to the curve y = f(x) at the point (x h yi), The numerical 
value of the derivative of fix) at (x h y{) is denoted by f(x{) or yi, 
and the tangent is a line through the point (x h yi) with the slope 
f(xi) or yi . Hence the equation of the tangent is 

(0 2/ — 2/i = /'Oi) (x - a*) or y -yi = yi(x - x x ). 

The equation of the normal at the point (x h yi) is readily found to be 

(n) x - x 1 = - f'(xi) (y - 2/1) or x - x x = -yi(y - yi). 

For example, we found that Dx s = 3 x 2 , and therefore the equations 
of tangent and normal to the curve y = x 3 at the point (x h y{) are 

y - y l = 3 x^(x - Xi) and x - x 1 = — 3 x^{y — y{). 

The point (2, 8) lies on this curve, and by substituting 2 for x 1} 
and 8 for y x in the last equations, we get for the equation of the 
tangent to this curve at the point (2, 8), 

^-8 = 3(4)0-2) or y - 12z + 16 = 0, 
and for the equation of the normal, 

z-2=-3(4)(j/-8) or 12^ + ^-98 = 0. 

24. Exercises. 

1. Find Dx 2 , and the equations of tangent and normal to the parabola 
y = x 2 at (xi, yi) and at (2, 4). 

2. Find D(3z 2 +2z— 1), and the equations of the tangents to the 
curve y=Sx 2 +2x— lat the points (1, 4) and ( — 1, 0). 

4 

3. Find D -, and the equations of the tangents to the equilateral 

x 

hyperbola xy = 4 at the points whose abscissas are —4 and 2. 

4. Find D and the equations of the tangent and normal to the 

x-\- 1 

curve y = at the point whose abscissa is 1. 

x-\- 1 

5. Find D(2 z 3 — 3 x 2 ) and the equations of the tangents and normals 
to the curve y= 2x z - 3 x 2 at the points whose abscissas are 0, ±1, I, 
and 2. 



CHAPTER IV 
RULES FOR DIFFERENTIATION 

25. Definition. To find the derivative of a function is to 
differentiate it, and the process itself is termed differentiation. 
We have now to establish certain formulae or rules for differentia- 
tion. In these formulae, c, n, p, q denote as usual arbitrary con- 
stants, while u, v, w, . . . denote functions of x. The increments 
of these functions corresponding to the increment Ax of x are 
Au, Av, Aw, ... In some cases we shall express the rule for 
differentiation in words as well as in symbols. These rules and 
formulae should be memorized. 

26. Differentiation of Polynomials. 

I. Dx = 1, The derivative of the argument is unity. 

Proof. Dx = lim -r— = 1 

J A*=oAx 

II. Dc = O. The derivative of a constant is zero. 

Ac 

Proof. Dc = lim — = 0, because Ac is always 0. 

J Ax=0AZ 

III. Dcxi = cDu; Dcoc = c. 

Proof. Dcu = lim -r — 
aj-o Az 

When the argument takes the value x + Ax, u takes the value u -f- 
Au, and cu takes the value c(u + Au)* Then the increment of cu 
is the difference between c(u + Au) and cu, that is 
Acu = c(u + Au) — cu— cAu; 

Dcu = lim ^ = cDu, . . . by Art. 20, 5, Cor. 

* See footnote on page 32. 
38 



§26 RULES FOR DIFFERENTIATION 39 

By placing u = x in this formula and taking account of I, we obtain 
the second part of the formula, 

Dcx = cDx = c. 
IV. D(± u±v±w ± . . . ) = zb Du ± Dv zb Dw ± . . . . 

The derivative of an algebraic sum of functions is the algebraic 
sum of their derivatives. 

D f r>/ , , . , \ v A(±u±v±w zb. .) 
Proof. D(db u do v ± wj zb . . . ) = lim — t > • 

When the argument takes the value x + Ax,* the algebraic sum 
zb ti zb t; zb w zb . . . takes the value ± (u + Aw) zb (t; + At;) ± 
(w + Aw) zb . . . .* Then for the increment of the algebraic 
sum we have 
A(±u±v±w± . . . ) = [±(u + Au) ± (v + Av) zb . . . ] 

- [±U±V± . . . ] 

= ± Aw ± At; zb Aw zb . . . . 
.*. D(±u±v±w± . . .) 

zb Au zb At; zb Aw zb . . . 



= lim 

Ax=0 AX 

= zb lim -T- zb lim -r- zb lim -r— zb . . . by Art. 20, 7 

A Z =0 AX Ax=oAx Ax=0 AX 

= zb Dw zb Dv =b Dw =b . . . . Q. E. D. 

„ , N _ ^ . ^ , X)t^ J>?e . Dv 

V. (a) Duv = vDu -\- uDv and ■ = — ■ -\ 

UV U V 

(b) Duvw = vwDu + wuDv + uvDw 

. Duvw Du . Dv , Dw 

and = ■ 1 • H 

UVW U V w 

(c) Duvw . . . ={vw . . . ) Du -\- (uw . . . )Dv 

+ (uv . . . ) Dw + . . . 

, Duvw . . . Du , Dv , Dw , 

and = h ■ — ■ + ■ b . • . . 

UVW ... U V w 



Proof of (a). Duv = lim — r — - 

Ai=0 AX 



Ax=0 

* See footnote on page 32. 



40 DIFFERENTIAL CALCULUS §26 

When the argument takes the value x + Ax,* u and v take the 
values u + Aw* and v + Ay,* and the product m> takes the value 
(u + Au)(v + At;). Then, for the increment of uv, we have 

Auv — (u + Aw) (y + At;) — uv = vAu + wAt; + AuAv. 

~ ,. vAw + wAv + AwAt> 
.*. Duv = hm t 

Ax=0 Ax 

y vAu .. wAt; . .. AwAt; , , , „ 

= hm -r— + hm -r h hm — : — , by Art. 20, y. 

ax=o Ax Ax - Ax ax=o Ax ' J 

Now, since u and t; are constants here (Au, Av, Ax are the variables) 

vAu 
it follows, from the corollary of <5, Art. 20, that the limits of -r — 

and -r— are t;Z>w and uDv respectively. As for the limit of — r — , 

,. AuAv .. Au,. A tat At; 

lim — t — = hm -t— hm Av or hm Au lim -r— 

Ax=0 &X a z =oAXaz=0 Ax=0 Aa;=0 &X 

= Du.O or O.Dt; = 0. 

Therefore finally, Duv = vDu + uDv. Q. E. D. 

Dividing this by uv we have the second form of the formula, 

Duv Du . Dv ~ ^ „ 
= — H Q. E. D. 

UV U V 

Proof of (6). Duvw = D(uv)w = wDuv + uvDw, by (a) 

= w(vDu + uDv) + uvDw, by (a) 

= vwDu + wuDv + uvDw. Q. E. D. 

Dividing by uvw we have the second form of the formula 

Duvw Du , Dv . Dw ~ ^ ^ 
= 1 1 Q. E. D. 

uvw uvw 

Proof of (c). This is an extension of (a) or (6) to the case of a 
product of any number of functions and may be derived by 
repeated application of the process by which (6) was derived from 
(a). Dividing the first form of (c) by uvw ... we get the second 
form. Q. E. D. 

These formulae may be expressed in words as follows: 
The derivative of a product of any finite number of functions is the 

* See footnote on page 32. 



§26 RULES FOR DIFFERENTIATION 41 

sum of all products that can be formed by replacing in the given prod- 
uct a factor by its derivative. 

The derivative of a function divided by the function is termed 
the logarithmic derivative* of the function, and the second form 
of (c) may be expressed in the following words : 

The logarithmic derivative of a product of any (finite) number of 
functions is the sum of their logarithmic derivatives. 

VI. T>u n = nu n ~^jyu; I>x n = noc"- 1 . 

Proof when n is a positive integer. 

First Proof. If in V (c) we put u = v = w = . . . , the prod- 
uct uvw . . . becomes u n , and we have 

Du n = u n ~ 1 Du + u n ~ 1 Du + u n ~ 1 Du + . . . to n terms. 

.*. Du n = nu n ~ l Du. Q. E. D. 

Au 11 

Second Proof. The direct proof. Du n = lim — 

Ax=o Ax 

By an argument similar to that used in the proofs of III, IV, and 

V(a), it may be shown that 

Au n = (u + Au) n — u n . 

Since n is a positive integer, (u + Au) n can be expanded by the 
binomial theorem, and therefore 

Au n = nu n ~ l Au + n ^ n ~ ^ u n - 2 (Au) 2 

. n(n — \)(n — 2) ... ._ , , ,. . 

+ — 23 u n ~\Auf + . . . + (Au) n . 

Dividing by Ax and making use of 7, Art. 20, there results 

^ „ ,. , Au . ,. n(n — 1) . Au . 

Du n = hm nu n ~ l - k f- lim - A — t z — - u n ~ 2 -r— Au 

a*=o Ax Ax ~o 2 Ax 

. v n(n — l)(n — 2) K „Am /a ,. , , _. &u, A s . 

+ hm-^ JA '- u n ~ 3 -r- (Au) 2 + . . . + hm -r- (Au)*- 1 . 

ax=o 2,.6 Ax Ax =o Ax 

Au 
Now by 5, Art. 20, lim nu n ~ l -r— = nu n ~ l Du, 

Az=0 AX 

*The significance of the term logarithmic derivative will appear later 
(Art. 69). 



42 DIFFERENTIAL CALCULUS §26 

and 

Um n(n-l) . ..(n-r + 1) u „_ r Au (Am) ,_ 1 = ^ 

Az=0 K ^ X 

for all positive integral values of r. 

Hence in the last expression for Dw n all the limits are except 
the first, and consequently 

Du n = nu n ~ l Du. Q. E. D. 

Placing u = x in this formula and applying I, we get the second 

part of the formula 

Dx n = nx n ~ l Q. E. D. 

These rules suffice for the differentiation of any polynomial or 
product of polynomials. 

Examples. 

1. To differentiate x 4 - 3 x 2 + 6 x + 8. 

Solution. D (x 4 - 3 x 2 + 6 x + 8) 

= Dx 4 -D3x 2 +D6x+D8 by IV 

= Dx 4 - 3Dx 2 + QDx by III and II 

= 4z 3 -6z+6 by VI and I 



2. To differentiate (2 x z + 3 x 2 - 5) 



Solution. D(2x z +3x 2 -5Y 

= 3(2x*+3x 2 -5) 2 D(2x 3 +3x 2 -5) by VI 

= 3 (2 x z + 3 x 2 - 5) 2 (2 Dx 3 + 3 Dx 2 ) . . by IV, III, and II 

= 3(2x s +3x 2 -5) 2 (6x 2 +6x) by VI 

= 18s(a;+l) (2x 3 +3x 2 -5" 2 . 

3. To find the derivative of (x 4 - 3) 2 {x 2 + 2) 4 . 

Solution. D(x 4 -3) 2 (x 2 +2) 4 

= (x 2 +2) 4 D(x 4 - 3) 2 + (x 4 - 3) 2 D{x 2 +2) 4 . . . .by V(a) 
= (z 2 +2) 4 2(z 4 -3)Z>0; 4 -3) + 

(a: 4 - 3) 2 4 (x 2 + 2)lD(z 2 + 2) by VI 

= (z 2 +2) 4 2(x 4 -3)4z 3 +(z 4 -3) 2 4 0c 2 +2) 3 2z 

by IV, VI, and II 
( = 8x(x 4 - 3) (x 2 + 2) 3 [(x 2 + 2) x 2 + (x 4 - 3)] 
' =8x(:r 4 -3) (z 2 +2) 3 (2z 4 +2a: 2 -3). 



§§27-28 RULES FOR DIFFERENTIATION 43 

Or we may first find the logarithmic derivative thus: 

D^-ZYJ^+ZY = D(*«-3)' fl(s»+2)« b V(fl) 2d fQrm 

(z"-3) 2 (z 2 +3) 4 (z 4 -3) 2 ^ (z 2 +2)* ' w ' 

= 2 -ff^f- 3 + i |^|^b y VI,IV,II,andVI 



= c r a; 2 1 1 8a:(2a: 4 +2a; 2 -3) 

L(a: 4 -3) a: 2 +2j (z 4 -3)(z 2 +2) 



Clearing of fractions, we get the same result as before. 

27. Exercises. Find the derivatives of the following functions : 

1. x*+x 2 +x-l. 7. (2z-3) 2 (3x-2) 3 . 

2. 2z 3 -4z 2 +:r. 8. (2a; 3 - 3) 2 (3a; 2 - 2) 3 . 

3. \x^-\x^-x. 9. (z+l) 2 (a;+3) 3 (x-5) 4 . 

4. (2a;+3) (3 a; +5). 10. w = (x+ 3)(z- 3) 2 (a;- l) 3 . 

5. (3w 2 -2)(2w 3 -3)- 11. 0(v)= (y+l)(y-l) 2 (y+3) 3 . 

6. (2w 2 -3^+7) 3 . 12. ?/= (x 3 -4) 5 (5a; 7 -12) 3 . 

28. Tangents and Normals to Curves. The rules for differen- 
tiation should be applied in finding the slopes of tangents and 
normals to curves. 

Examples. 

1. Find the equations of the tangent and normal to the curve 

y = x z - 2 x 2 + 4 at the point (1, 3). 
Solution. We find Dy = D(x z - 2 a; 2 + 4) = 3 x 2 - 4 x. 
When x=l, D?/=3-4=-l. 

Writing then the equation of a line through the point (1, 3) with 
the slope —1, we have as the equation of the tangent line 

y— 3 = — (x — 1) ovy+x— 4=0. 

Writing the equation of a line through (1,3) with the slope 1, we have the 
equation of the normal 

y— 3= x— 1 or y— x— 2=0. 

2. Find the equation of the tangent to the ellipse 

b 2 x 2 + a 2 y 2 = a 2 b 2 at the point (x h yj. 
Solution. We have 

Db 2 x 2 +Da 2 y 2 = Da 2 b 2 = by IV and II 

or 2b 2 x+2a 2 yDy= 0.. by III and VI 

y'=Dy=-— and j// = - — . 

a 2 ?/ a 2 ?/! 



44 DIFFERENTIAL CALCULUS §§29-30 

Hence the equation of the tangent is 

which is readily reduced to the usual form, 

b 2 XiX + o}y x y = o?b 2 . 

3. Find the equations of the tangent to the curve xy 2 + x 2 y = 6 at the 
point (1, 2). 

Solution. D(xy 2 + x 2 y) = Dxy 2 + Dx 2 y = D 6 = by IV and II 

xDy 2 + y 2 Dx+x 2 Dy+ yDx 2 =0 by V(a) 

2 w'+ y 2 + x 2 y' + 2xy=0 by VI and I 

(2xy+x 2 )y'=-(y 2 +2xy), 

whence ' y' = — %-^ %• 

u 2xy+x 2 

When x = 1 and y=2,y'=—§, and hence the equation of the tangent is 

2/ _2=-l(x-l) or 5?/+ 8a:- 18=0. 

29. Exercises. 

Find the equations of the tangents and normals to the following curves 
at the points indicated : 

1. y = x 2 -\- 2x at (x h y{) and ( — 1, —1). 

2. y~ x 3 at (0, 0), (1, 1), and (-2, -8). 

3. y=x\x 2 - 1) at (0, 0) and (±1, 0). 

4. y 2 =2mx at (xi, yj. 

5. x 2 - y 2 = 1 at (^ yj and (2, ±V3). 

6. 2y 2 =x 3 Sit (2, ±2). 

7. ?/ 3 =2a: 2 at (±2,2). 

8. z?/ 3 = lat (1, 1) and (-1,-1). 

9. x z y= 8 at (±2, ±1) and (±1, ±8). 

10. 2/=^at(l, 0), (2,-i)and(-2, 1). 

x l 

11. x z — 3axy+ y z = at (xi, yi). 

12. 2x 2 y 2 +x-y=2&t (1, 1) and (-1, -1). 

30. Differentiation of Fractions. 

u vDu — nDv 

VII. JD - = s 

v ir 



§31 RULES FOR DIFFERENTIATION 45 

The derivative of a fraction is the product of denominator by 
derivative of numerator minus the product of numerator by deriva- 
tive of denominator, all divided by the square of the denominator. 

A- 

Proof. D — = lim -j — • 
v az=o Ax 

When the argument takes the value x + Ax, the fraction takes 
the value . , and consequently 

A u _ u + Au u _ vAu — uAv 
v v + Av v v(v + Av) 

Au Av 

^u ,. Ax Ax vDu — uDv - _ ^ 

D - = hm — ? — , A , = = Q. E. D. 

v ax=o v\v + Ay) y 2 ^ 

1 Dv 1 1 

v v* 5 oc or 

The first formula is gotten by setting u = 1 in VII, and the second 
by setting v = x in the first. 

31. Proof of VI when n is a Negative Integer. 

Let n = —m, m being a positive integer. 
We are to prove that Du~ m = — mu~ m ~ l Du. 

By VIII. Du~ m = D\=-±~. 

Since m is a positive integer, Du m = mu m ~ 1 Du 

Therefore, Du~ m = - u2m = -mu~ m ^ Du. Q. E. D. 

Examples. 

1. To find the derivative of - — : — - • 

4: X S - 1 

Solution. 

3a 2 + 2 _ (4a; 3 - 1) £>(3a; 2 + 2) - (3a; 2 + 2) D(4g»- 1) , VTT 

^4a; 3 -l (4a; 3 - l) 2 yV11 

= (4a; 3 -l)6a;- (3a: 2 + 2) 12 a; 2 = -6x(2x*+ 4s + 1) 
(4a; 3 - l) 2 (4a; 3 - l) 2 

2. To find the derivative of t^~ • 

(a; 2 +2) 3 



46 DIFFERENTIAL CALCULUS §§32-33 

Solution. 

D (X*+ l) 2 = (X*+ 2) 3 Z>(z 2 + 1)2- ( X 2 + l)2 Z )( g 2 + 2)3 

(z 2 +2) 3 (x 2 +2) 6 y 

_ (x 2 + 2)X(o; 2 + 1) 4 a; - (s 2 + l) 2 jbH^ 6 a; 

(z 2 +2)* 4 
_ 2z(:c 2 +l)(l-:c 2 ) 
(z 2 +2)4 

3. To find the derivative of — — — • 

(x 3 — 4) 2 

A'rsJ Solution. 
D ( -^T4Ji = ^'-4)- 2 = -2(^-4)-'3^= ( -^j by VI 

Second Solution. 

p 1 = ~ Z) ^ 3 - 4 ) 2 (by VIII) = -2(^^3o; 2 = -6* 2 . 
(z 3 - 4) 2 (z 3 - 4) 4 ; * ; (z 3 - 4)* 3 (z 3 - 4) 3 

32. Exercises. Find the derivatives of the following frac- 
tions : 

1. £±1. 6. Ml 3 - 11.-^-. 16. 



z 2 +l x*-3 x 5 -l (x-2Y 

2 1 ~ x 7 t+A. 12 ^ 2 -3x+l 17 x 3 





x 3 -3 


7. 


z 2 +l 


z 3 +l 


8. 


4 


3-z 2 


9. 


x 3 
x 4 +l 


r» 


x 



z 2 +l x*+l x*+x 2 (z 2 +l) 2 

3. In?. 8 . * 13. * + *+? 18. 



x 2 +2 3-z 2 z 3 -3x 2 (z 2 -l) 2 

4 * q __*!_. 14 * 3 ~* in (s+1) 2 

" (3z 3 +l) 3 * z 4 +l ' z 4 +3a; 2 -i " 2z 3 +l 

2^+5 _j L _. = a 2 20 (3z 2 +l) 2 

z 3 -3 z 5 +l (2z+l) 3 (z 2 +2) 3 

(2z 2 +D 3 22 (s 5 -2) 3 
' (3z 2 -2) 2 * (z 3 +l) 5 

33. Differentiation of Irrational Functions. 

Proof of VI when n is a fraction. 

We are to prove Du q = — u 9 Du, where p and q are integers. 



Since q is an integer, we have by VI, 

/ p\q I p\q-i v p _p v 

D\u q ) = q\u q ) Du q = qu q Du q . 



§33 RULES FOR DIFFERENTIATION 47 



Also 


D\u«) = 


Du p = pu 


P~ l Du, 


because p 


is an integer. 






Hence 


qu 


p— ? - 
lDu q = 


■■ pu p ~ 1 Du, 


and 




p 

Du« = 


- it* Du. 
0. 



Q. E. D. 

Formula VI has now been proved for all rational values of n. 
Examples. 

^2 2 2 -l 2-1 9 

1. Dx* = -x* =^3 



3 3 3 \Z X . 

3. D VH 7 ^ 2 = #(1 + x 2 ) 2 = 1 (1 + a*)i-\D(l + x 2 ) = J (1 + x 2 )- ^2 x 



VI + x 2 
*• "Vl-x U \l-x) 3U-J 1-s 3Vl+z] (1- 



3(l + x)i(l-x)* 



0. i> ~ = U r = 

1 - </x 1 - X3 (1 - X^) 2 



3zf (1-a^) 2 
A second method of solving 3 and 4 is as follows : 
3. Let y = Vl + x 2 ; then if = 1 + x 2 , 2 y Dy = 2 x, 
Dy=-=* 



V Vl + r 



k /^;then,3 = ^, 3 , 2 7), = ^ 
2 2 



#2/ 



32/ 2 (l-a;) 2 3(l + z)*(l- *)* 
This method could of course be applied to the solution of 1 
and 2, but there would be no advantage in using it in those 
problems. 



48 DIFFERENTIAL CALCULUS 



34-35 



34. Exercises. Determine the following derivatives, regarding 
x as the argument in each case: 



1. D(2V~x+iy. 6. Z)(af-xl)f. 1Q D VT^Z 

2. D V2s 2 +.l. 7. d u • ' y3 

3. D(l-2ifll Vl+U ii. £)1±V^ 
8. Da; Vl -x\ 1 - Vz 



4. D(l-2^f) 3 . 

5. D Vl - Vy. 



9. D^szJ. is. z)^. 



13. D— ; Z)^±^. 21. D^±-?. 



14 - D ^ ;Z) ^)- 22. lM=Jte 



vx+u 

f(x) + g(x) 

V t> 2.V^ ' 23. Z) V !f V 7 

Vu+ vi; 
In the three following exercises 

test results by substituting in 15 



16. 


°v;::- 


17. 


Wm- 


18. 


W9- 


19. 


vx 3 V 6 X 


20. 


D u 2 + x 



24. D 



Vuv(Vu+ Vvf 
Va+x — Va — x 



Va + x + Va — x 

Hint. Rationalize the denomi- 
nator before differentiating. Solve 
also by substituting in exercise 23. 

25. D vir^+vr^. 

Vl + x 2 - Vl - X 2 
x 2 +u ' 26. D Vu 2 x+ux 2 . 

35. A More Descriptive Notation for the Derivative. 

Such a formula as 

Du n = nu n ~ 1 Du 

does not tell us what the argument is. When it is desired to bring 

the argument into evidence, this may be done by writing the 

argument as a subscript of the derivative symbol. Thus, u being 

a function of x, we may denote its derivative by D x u, which is read 

"the derivative of u as to x." With this notation the foregoing 

formula becomes 

D x u n '— nu n ~ l D x u. 



§§36-37 RULES FOR DIFFERENTIATION 49 

These two formulae mean of course the same thing. Their sole 
difference is that the second tells us explicitly that x is the argu- 
ment, while the first does not. If u itself be the argument, the 
formula becomes 

D u u n = nu n ~ l D u u = nu n ~ l , 

which is obviously the exact equivalent of Dx n = nx n ~ l . 

There is often no need to indicate the argument by the notation. 
Thus in the exercises of the preceding pages, x was assumed to 
be the argument at the start, and so the subscript notation was 
unnecessary. 

36. Exercises. Perform the following differentiations: 

1. D x uv; D u uv; D v -; D x iix. g j) — * x 

■vVz+1 



v 



2. D u *±±- 



!. 2>V ; 



z 3 +l V3 

8. D 



4 ^(TwIF (2 * 2+1)i 



10. D 



(l + rf) 1 
+ 1 

2 +: 

z 3 +4)f 



9. D ^ 3+1 ^ 



5. D t Vx+Vl+x 2 . (3z 2 +4)t 

37. Differentiation of Implicit Functions. When u and x are 
connected by an equation which implies that u is a function of x 
without giving expressly the form of u in terms of x, u is said to be 
an implicit function of x. Thus, the equation 

u 2 -2 ux + x 2 -3z + l=0 

gives u as an implicit function of x (or x as an implicit function of 
u). Solving for u, we get u = x ± V3 x — 1, which gives u as an 
explicit function of x. It is not always convenient, and is some- 
times impossible, to solve such an equation, and so to express the 
function explicitly in terms of the argument. For example, 

2 u h - 5 x 2 u 2 + 2 x b - 3 = 



50 DIFFERENTIAL CALCULUS §38 

cannot be solved for u. Nevertheless it is quite a simple matter 

to determine D x u. Thus, 

D s (2u* -5xV + 2z 5 - 3) = 0, 
whence 

lOwV - 10 x 2 uu' - 10 zw 2 + 10z 4 = 0, 
or 



xu 



2 _ 



(u 4 — x 2 u)u' = xu 2 — x 4 , and v! = 

w* — x*u 

From a study of this example the general principle involved in the 

process becomes evident, and may be expressed thus: 

Ifu be defined as an implicit function ofxby the equation 

f(u, x) = 0, then D x f(u, x) + D u f(u, x) D x u = 0, 

D x f(u, x) 

and D x u =— ■^ F — L ?. r- 

D u f{u, x) 

A proof of this formula is given in Art. 203, formulae I& and /&'. 

We have already had, in Arts. 28 and 29, a few examples of finding 

derivatives of implicit functions. We add here additional exercises. 

38. Exercises. 

1. Find u' from the equation (v? + x 2 ) 2 - 4 ux + 3 = 0. 

2. Find u' from the equation v?x + ux 3 — 4 = 0. 

3. * Find D$u from the equation 4 u z + 3 ud 2 + 6 0= 4, and determine 
the value of Dqu when u= —\. 

4. Find Z> u from the equation of exercise 3, and determine its value 
when u= —\. 

5. Find the equations of tangents and normals to the curve 

if — 3 axy + x z = at the points (I a, fa), (1 a, fa). 

6. Find the equations of tangents and normals to the astroid 
£3-f-?/3 =a3 at the points where y = x, and also at the points where the 
curve cuts the axes. 

7. Find y' from the equation (x+ y) 4 = ax 3 both by solving for y and 
without doing so. 

8. Find the equation of the tangent to the curve of exercise 7 at the 

point £ci,2/i),and show that this tangent passes through thepointf -^ —^ J. 

9. Find the equations of the tangents to the curve y 2 = a\/x 2 '-\-y 2 at 
the points (0, ±o), (fa, f V3 a). 

10. Find the equations of the tangents to the curve y 2 = 5 x z — 3 x 5 at 
the points whose abscissas are 0, 1, Vf. 



§39 RULES FOR DIFFERENTIATION 51 

39. Differentiation of Trigonometric Functions. In the proofs 
that jfollow we shall make frequent use of the principles of limits, 
a . . . yj, Art. 20; but for sake of brevity we shall omit specific 
reference to them. As heretofore, w is regarded as a function of x, 
and the second part of each formula is obtained by placing u — x 
in the first part, and observing that Dx = D x x = 1. 

IX. D sin u = cos uDn = sin I u -j- „ ) Du; 

D since = cos cc= sin fa? + ^-j- 

D t r> • r Asinw 
Proof. Dsmu = hm — x 

Az=0 AX 

Now, A sin u = sin (u + Aw) — sin u 

= sin m cos Aw + cos w sin Aw — sin w 
= cos w sin Aw — sin w(l — cos Aw). 
Therefore, 

A sin w sin Aw Aw . 1 — cos Aw Aw 

— 7 = cos w — t • -r sin w 7 • -t— • 

Ax Aw Ax Aw Ax 

In determining the limits we make use of the results obtained in 

Art. 11, examples 3 and 4, viz., 

,. sin Aw . , ,. 1 — cos Aw 

hm — 7 = 1 and hm 7 = 0. 

Au=0 AU Au=0 AU 

Hence lim — t = coswDw, 

a*=o Ax 



and 



D sin w = cos wDw = sin ( w + « ) Du. Q. E. D. 



X. D cos u = — sin uJDu = cos (u +— J j>^ ; 



Dcosoc = — sin a; = cos fa? + 5) 



7T 

Proof. Writing ^ — w for w in IX, 



we 



have, D sinf^ — wj= cosf = — u)d( ■= — wj, 



52 DIFFERENTIAL CALCULUS §39 

whence 

D cos u = —sin uDu = costu + ^jDu. Q. E. D. 

Problem. Prove this formula by "the"direct method used in the proof 
of IX. 

XI, J>tanw = sec 2 uDu; 2>tana? = sec 2 a% 

sin u cos u D sin u — sin u D cos u 



Proof. D tan u = D 



cos u cos 2 u 

(cos 2 w + sin 2 u)Du 



cos 2 w 



by IX and X 



D tan w = sec 2 uDu. Q. E. D. 

XII, D cot u = — esc 2 uDu; D cot a? = — esc 2 a% 

Proof. This is proved by writing - — u for u in XI, or by 

cos w 

differentiating •, or by differentiating Let the student 

sm u tan u 

give each of these proofs. 

XIII, D sec u = sec u tan uDu; D sec a? = sec a? tan a?. 

Proof. Let the student prove this by differentiating • 

cos w 

XIV, D esc t* = — csc u cot uDu; D esc a? = — esc a? cot oc, 

IT 

Proof. Let the student prove this either by writing - — u for 

u in XIV, or by differentiating — 

sin it 

Examples. 

1. D tan 4 x = sec 2 4 #D 4 z = 4 sec 2 4 z. 

2. D sin 2x z = cos 2 z 3 Z) 2 z 3 = 6 z 2 cos 2 z 3 . 

3. Z) sec 3 5 x 2 = 3 sec 2 5 z 2 D sec 5 x 2 = 3 sec 2 5 x 2 sec 5 z 2 tan 5 x 2 D 5 z 2 

= 30 x sec 3 5 z 2 tan 5 x 2 . 

4. Da; cot VI +x 2 = cot Vl + z 2 - a; csc 2 Vl +z 2 D Vl + z 2 

, x 2 csc 2 Vl +z 2 

= cot VT + z 2 7== 

Vl + z 2 



§§40-41 RULES FOR DIFFERENTIATION 53 

40. Exercises. Perform the following differentiations: 

1. D tan 3 2 x. 11. D (x sin x+ cos x); 

2. D(x ± cos 2 a;). Z>(sin x - x cos a;). 

3. D{6- sine cose). 12. D t esc a cot a. 

4. D.ri nV^ 13 ' ^xBin(ii+e)coB(ti- g ). 

5. D.Vl + sin^^Vl-cosy." 14. D y J l + SmU ' 

6. 2)(tan2-s);D(cot2+s). V lsmw 

7. D sin 2 x cos 3 a\ 15. D y \ — 

v \ 1 + cos w 

8. Z)(sin y - £ sin 3 ?/). 16 D ^ tan3 w _ tan u + w ). 

9. D(tan?/+ i tan 3 ?/). 17. D(_cosz + f cos 3 x- icos 5 x). 
10. D Vsec 2 2 ; D esc 3 3 3 . 18. D(sec x - I sec 3 z+ I sec 5 x). 

Differentiate both members of each of the following equations : 

19. sin 2 x = 2 sin z cos x. 00 o - 1 ~ tan 2 x 

zz. cos z x- ! , t 2r * 

20. cos 2 a; = cos 2 x — sm 2 x. 9 

21. cos 2 x = 1 - 2 sin 2 x. 23> tan 2 * = i+tan 2 z' 

41. Differentiation of Circular Functions. In the following 
the student should carry out all the proofs outlined. The second 
part of each formula is~obtained by writing x for u in the first 
part and noting that Dx = D x x = 1. 

! Dw . ■, 1 
XV. J>sin~ x ^= ; =: z>sm _1 a? = 



Proof. Let ?/ = sin -1 w, whence sin y — u. 
Differentiating this equation, we have 

Du 



cosyDy = Z)w, Dy = 



cos ?/' 



and, since Dy = Z) sin -1 u and cos 2/ = V 1 — u 2 , 

D sm-'u = y Du Q. E. D. 

^„r ^ 1 Du . 1 

XVI. £>C0S~ 1 u = ; DC0S -1 a?= — 



Vi-™ 2 Vi-cc £ 



54 DIFFERENTIAL CALCULUS §41 

Proof. This may be proved by the method used in the proof 
of XV, but may be more readily obtained by differentiating the 
identity 

_i *" • -i 

cos 1 u = - — sin -1 u. 

A 

XVII. Dtan- 1 u = - : W z ; Dta.n- 1 sc 



l + u?' ' 1+x 2 

Proof. Let y = taji~ 1 u, whence tan?/ = u. Differentiating 
this equation, we have 

sec 2 y Dy = Du, Dy = — — • 
sec 2 y 

But Dy = D tan -1 u and sec 2 y = 1 + u 2 , 

Du 
^tan- 1 ^^— — -• Q. E. D. 

1 + u 2 ^ 

XVIII. DC0t- 1 u=- . ■ U „ • DC0t- 1 sc=- _ t 1 - , 

1 + U 2 ' 1 + K 2 

Proof. This may be proved as XV and XVII were proved, or 
better, by differentiating the identity, 

cot -1 u = - — tan -1 u. 

Du , 1 

xix. d sec - x u = ■ ; d sec- 1 x 



u 



Vu 2 - 1' kVk 2 -1 



Proof. This may be proved as XV and XVII were proved, or 
by differentiating either one of the identities, 

sec -1 u = cos -1 - or sec -1 u = tan -1 Vu 2 — 1. 

u 

Du , 1 

XX. DCSC~ 1 U= ; J>CSC -1 x= — 



u Vu 2 — 1 x Vx 2 — 1 

This may be proved by the method used in the proofs of XV 
and XVII, or by differentiating any one of the identities, 

esc -1 u = - — sec -1 u, esc -1 u = sin -1 — , or 

csc -1 ti= cot -1 Vu 2 — 1, 



§42 RULES FOR DIFFERENTIATION 55 

In strict logic the double sign ± should be written before each ■ 
of the radicals in the foregoing formulae. There is consequently 
an ambiguity in such of the formulae as contain radicals. Let 
the student show that the ambiguity is due to the fact that the 
circular functions are infinitely many- valued functions of their argu- 
ments (see Art. 2). 

Examples. 

!. DBil ,-.2£Z-L. S_... 1 



vA-m 



V2 + X-X 2 



2. Z>.tan-*A^- 2v^_ = _4 j£! = _ -D& 

2Vu 






1 



3. Z>sec-Vl— ^ DVuj_ ^.2^±i mt 1 

Vi + zVi+z-i Vi + tVt 2(i + t)Vt 

42. Exercises. Perform the following differentiations: 

1. D sin- 1 2x. 8. D(x Vl - x 2 + sin- 1 ^). 

2. D.ooB-^l-t*). 9 Dtan , t 1 . 

3. ^tan-l- V?+2t 

a n i /- D C ot-ii±^^ 

4. Usee -1 vz. sinx 

5. Dsin- 1 ^- 10. Df^fi+sec-^V 

cc + 4 V a; 2 / 

6. jDcsc- 1 ^- n - ^T-f^+tan- 1 ^). 

x—4c \l + x 2 / 

7. D(aBin-i?-V5^). 12. s(f^g+ 3 ton-^j! 

13. 7)sin-^ 1 = 7)cos- 1 ?^= J Dtan- 1 ^^. 

3+1 »+l 2 vV 

14. Z)sin-^-±i = Z) S ec- 1 ^±^ = Dtan- 1 -^L. 

x + z V2z+3 V2X+3 

15. /)sin- 1 4^=^csc-^^+- > )=Z)tan- 1 -^- 

1 + z 2 2\ xj 1-x 2 



CHAPTER V 

DERIVATIVES OF HIGHER ORDERS 

43. Definitions and Notation. Since Df(x) is in general a 
function of x, it may itself be differentiated. The derivative of 
Df(x) is termed the second derivative of f(x), or the derivative of f(x) 
of second order, and is denoted by D 2 f(x), so that 

D 2 f(x)^DDf(x). 

D 2 f(x) has in turn a derivative termed the third derivative of f{x), 

or the derivative of f(x) of third order, and is denoted by D 3 f(x), so 

that 

D*f(x) = DD 2 f(x) = DDDf(x). 

In like manner we get the fourth, fifth, nth derivatives of f(x), 

which are denoted by 

D*f(x), D*f(x), . . . D»f(x). 

When it is necessary to distinguish it from higher derivatives, 

Df(x) is termed the first derivative of f{x), or the derivative of the first 

order. Other symbols for the several derivatives are 

fix), /"(*), /'"(*), m . . . fix). 

If y be the functional symbol, the several derivatives of y are de- 
noted by 

Dy,D 2 y,D*y, . . . , or by y f , y" , y'" , . . . , or by^ 1 ), y^,y^. . .. 
The process of finding successive derivatives is termed successive 
differentiation. 

44. Successive Differentiation of Explicit Functions. 

Examples. 

1. Let f(x) = x A -3x z +5x 2 - 7x+3. 

Then f(x) = 4z 3 - 9z 2 + 10z- 7, 

f\x)= 12 x 2 - 18 x + 10, 
f"(x) = 24:x- 18, 
f"(x) = 24, 
and all higher derivatives are zero. 

56 



45 DERIVATIVES OF HIGHER ORDERS 57 



2. Let y 



1 + x 2 

Then '"aTST y = D (iT^ = 7T7^' 

, // = D 2x(x 2 -3) __„ x 4 -Qx 2 +l . 
V (1 + z 2 ) 3 (1 + x 2 ) 4 ' 

tv nr ^^ 4 - 6x 2 +l) -i 24x(a; 4 -10x 2 +5 ) 

w IV = D L~ 6 d+x 2 ) 4 _r a+x 2 ) 5 "' 

etc. etc. etc. 

45. Successive Differentiation of Implicit Functions. 

Examples. 

1. Let us find y-', y", y'" from the equation of the ellipse 

a 2 y 2 -\-b 2 x 2 =a 2 b 2 . 
First Solution. Differentiating as in Art. 28, example 2, we have 

a 2 yy' + b 2 x=0. (1) 

Differentiating this equation, 

a 2 yy" + a 2 (y'y + b 2 =0. (2) 

Differentiating again, 

a 2 yy'" + a 2 y'y" + 2 ahj'y' ' = 0, 

or yy"' + Zy'y" = 0. (3) 

From (1) we get 

b 2 x 

y = — t ■ 

a?y 
Substituting this value of y' in (2), there results 

ayy" + b 2 (b 2 x 2 +a 2 y 2 ) = 0, 
which, by aid of the given equation, reduces to 

6 4 
cPyty" + b 4 = 0, whence y" = — — — • 

a z y 6 

Substituting in (3) the values of y' and y" just found, we have 

^"'+!£? =0 ' whence y'" = - z -^- 

b 2 x 
Second Solution. We obtain y' as in the first solution, y' = — — - • 

a?y 

This we differentiate and get 

v "_ & D x_ b 2 y-xy' 



58 DIFFERENTIAL CALCULUS 



b 2 x 
Substituting for y' its value — — -, we have 

a 2 y 



b 2 x 2 
b* y+ a 2 y b 2 a 2 y 2 + b 2 x 2 
a 2 y 2 a 4 y 3 


_ 6 2 a^ 
a 4 y 3 


Hence y" = — —— • 
a-y 3 




Differentiating this last equation, we get 





„,_b* 3y 2 y' _ 3 5 V _ 3 fr 4 / b 2 x \ _ 3b G x 
V a 2 y« ~~ a 2 y l a 2 [ a 2 y 5 ) a,y 

No directions can be given for determining beforehand in any given 
case which of these methods will be the simpler. 

There is a third method that could be used in the problem before us, 
viz., that of solving the given equation for y and then differentiating 

successively the result, y = ± - Va 2 — x 2 . This third method would not 

be applicable if the given equation could not be solved for y, as is the 
case in the following example. 

2. Let us determine y' and y" from the equation 

y z — 3 axy + x 3 = 0. (1) 

Differentiating this equation, we have 

Sy 2 y' - 3 axy' - 3 ay + 3z 2 = 0, 
whence 

(y 2 — ax)y' — ay + x 2 = 0. (2) 

Differentiating again, 

(y 2 - ax)y" + y'{2 yy' - a) - ay' + 2 x = 0, 
and {y 2 -ax)y" -2ay' + 2yy' 2 +2x = 0. (3) 



From (2) we get y' = -f , and on substituting in (3) and reducing 

y l — ax 

there results 

(y 2 — axfy" = — 2 xy(a 3 + x z — 3 axy + y z ) • 

Then, by virtue of (1), 

= ~ 2 a 3 xy t 
y (y 2 - ax) 3 

46. Exercises. 

1. Find all the derivatives of 

4 x h - 5 x* - 10 x 2 and (x - l) 3 (x + 2) 2 . 



§46 DERIVATIVES OF HIGHER ORDERS 59 

2. Iiy=~r,Zndy',y'',y''',y™. 

jr- , 11/ 2,1, 

Hint. y=---,y = - -\ + -r, etc. 

X 2 X X Z X 2 

t 2 

3. If u = - — r , find u' and u" . 

f+1' 

4. Find the first four derivatives of \/x. 

5. Deduce a general formula for the nth derivative of x 8 , and apply it 
to the solution of Ex. 4. 

1 (-l) n \n 



6. Show that D n 

x x"^ 

7. Show that D n sin x = sin 



(*+»=)• 
(•+•!) 



8. Show that D n cos z = cos 

9. Find D 2 - 9 * • 

0+2) 2 

10. Differentiate the equation ?/ 2 = a: 2 + 1 five times and show that 

v __ 15z(4:r. 2 -3) 

11. If y z + x 3 = 1, show that ?/"' = -2 1 + ^ 3 ■ 

12. Find y' and y" from the equation £3 + y* = as . 

13. Differentiate six times the equation y 2 = x z — 3 x. 

14. From the equation u = tan we get 

u' = sec 2 = 1 + u 2 , \ u" = uu'. 
Continue the differentiation and show that 
u™= 8u(l + u 2 )(2 + 3u 2 ). 

15. If u = sec 0, show that u" = 2 u z — u. 

Continue the differentiation and show that ^ VI = 61 when 0=0. 



CHAPTER VI 

SOME GENERAL PROPERTIES OF FUNCTIONS: CON- 
VEXITY, CONCAVITY, AND FLEXES 

47. The Interval. It is often desirable to consider the be- 
havior of a function, not for all values of its argument, but for 
values within an interval. We fix upon the z-axis two points with 
abscissas a and b, and call the segment lying between them "the 
interval a to b" or "the interval a = x=b." We may then 

confine our attention to those 
values of the function which lie 
within this interval. When the 
graph is drawn as in the figure, 
we may speak of the behavior 
of the function "along the arc 
AB" instead of within the in- 
terval a ^ x = b. The interval 
may be very small, or it may be 

large enough to contain the entire graph. We may even have the 

interval — oo = x = + oo . 

48. Continuity. Confining ourselves to real values of functions 
(it is only for real values that a function has a graph), we define a 
function to be continuous within an interval when its graph is un- 
broken within that interval, and to have a discontinuity wherever 
there is a break in its graph* 

In Art. 8 we virtually defined a discontinuity as a point at which 

* Note to the Teacher. — This geometrical definition and discussion of conti- 
nuity is lacking in rigor, in that it appeals to the geometrical intuition, but it 
suffices for the purposes of an elementary course, and wall be found to appeal 
more strongly to the beginning student than would a more rigorous treatment. 
A rigorous analytical definition is given at the end of this article. 

60 




§48 



SOME GENERAL PROPERTIES OF FUNCTIONS 



61 



the function becomes nonexistent. At such a point there is a 
break in the graph and the discontinuity falls under our present 
definition, but we shall now give a couple of examples showing 
that a break in the graph may occur even when the function does 
not cease to exist there, and that consequently our present defini- 
tion is more comprehensive than the former one. 



Example 1. Let the function G(x) mean " the greatest whole number 
in x" so that when -1< x < 1, G(x) = 0; when l^x < 2, G(x) = 1 ; when 
2 ^ x < 3, G(x) = 2; . . . when 
- 2 < x ^ - 1, G(x) = -1; and in 
general n being a positive integer, 
when n^x<n-\-l, G(x) = n, and 
when — (n+ 1) < x= - n, G(x) = 
-n. The graph of G{x) consists 
of a series of discrete line seg- 
ments parallel to OX, and of 
length 1. The function has dis- 
continuities, breaks in its graph, 
at the points 

x = ±l, ±2, ±3, . . . 



(-1,0) O 



(-2,-i; (-1.-D 

(-3,-2) (-2,-2) 
(-4,-3TT-3,-3) 



( M) (5 ,4) 
(3. 3) (1 .3) 
(2, 2) (3 .2) 
(1, 1) (2 .1) 

(1,0) . v 



but is nowhere nonexistent. As x 

passes from a value a little less 

than n to a value a little greater, G(x) makes a spring from the value 

n — 1 to the value n. 



Example 2. A portion of the graph of the rather complicated function 

4 



fix) 



2 x ~ 2 +x 



is shown in the figure. This function is nowhere nonexistent, but has a 
discontinuity at x = 2, where the function springs from the value +1 to 
the value — 1. For when x approaches 2 from the left (i.e., remains 

l 



3-2 



always less than 2), the exponent of 2 X remains always negative and 

l 
—s 4 

2 decreases without limit, and so f(x) approaches the value - — 1, or 1. 

A 

On the other hand, if x approaches 2 from the right, the exponent of 



62 



DIFFERENTIAL CALCULUS 



§48 



2 remains positive and 2 X 2 increases without limit, and j(x) ap- 



i_ 

x — 



proaches the value — — 1, or — 1. When x = 2, 2 may be regarded 

as nonexistent, but f(x) exists, and has the value +1. The student 

should calculate a few values of 
f(x) by the aid of a table of loga- 
rithms, say for x = 1.7, 1.8, 1.9, 2.3, 
2.2, 2.1. 

Discontinuities like those just 
described will not be considered 
further because very few of the 
functions dealt with in this book 
have such discontinuities. The 
functions to be met with in the 
following pages are for the most 
part such as are continuous 
throughout their whole extent, 
or such as become discontinuous 

through taking the forms ~ or oo, 




like 



smz 



,2 _ 



and tan x. 



x x — a 
Observe that a function nei- 
ther ceases to exist nor becomes 
discontinuous merely by taking complex values. For example, 
Vx 2 — 1 is imaginary for all values of x between —1 and +1, but 
is nowhere nonexistent or discontinuous. 

The rigorous analytical definition of continuity is as follows: 
If f(a) exists, and if lim f{x) = f(a), no matter in what way x 

x=a 

approaches its limit a, f(x) is defined to be continuous at a. 

If, on the other hand, f(a) does not exist, or if, when f(d) does exist 
x can be made to approach a in such a way that lim f(x) y£ f{a), f(x) 

x=a 

is defined to be discontinuous at a. 

It can be shown that every case of discontinuity thus far men- 
tioned is covered by this definition. 



§§49-50 SOME GENERAL PROPERTIES OF FUNCTIONS 



63 



49. Discontinuities of f'(x). A function may itself be con- 
tinuous within an interval and yet have a derivative which is 
discontinuous within that interval. Such are the functions whose 
graphs are shown in the accompanying figures. In the first figure 
the derivative is discontinuous at A, B, and C, because at each of 





P X 



these points the tangent is _L to OX and consequently f(x) is oo . 
In the second figure the derivative is discontinuous at D, for at 
that point the tangent springs from the position T + to the position 
T~, and f'(x) springs from the value tana to the value tan/3. 
f'(x) is here discontinuous like the functions in Art. 48. Points 
such as D are termed angular points. 

50. How a Function Changes Sign. When a function changes 
sign its graph passes from one side of the x-axis to the other side. 
Now it is geometrically evident that a curve can pass from one 
side of OX to the other side in one of the following ways only. 
(See figures on next page.) 

(a) By crossing OX as in Fig. (a). 

(6) By a leap or spring as in the case of the curve of example 2, 
Art. 48. Fig. (b). 

(c) By " passing through infinity" as in Fig. (c). This is the 
case of the tangent curve y = tan x within any interval that con- 

7T 7T St 3x , 

tarns a;=^or — « > or — > or — — > etc., etc. 

(d) A curve may, as in Fig. (d), lie on both sides of OX within 
an interval because of having two or more branches within the 
interval, one lying on one side of OX, the other on the other side, 
while the curve itself within the interval does not cross OX, nor 
leap across it, nor pass around through infinity. Such a curve 



64 



DIFFERENTIAL CALCULUS 



50 



is of course the graph of a two- or more-valued function. In each 
of the cases (6) and (c) the curve has a discontinuity. It is plain, 
then, that a curve which consists of but a single branch and has no 




Eig.(c) 



Fig.(d) 



discontinuities can pass from one side of the #-axis to the other 
side only as in (a), by actually crossing that axis. Such a curve is 
the graph of a real single-valued and continuous function. We 
have, then, 

Theorem 1. A real, single-valued, and continuous function which 
has both + and — values within an interval must be zero at least once 
within that interval. 

Or, 

A real, single-valued, and continuous function can change sign only 
by passing through zero. 

The following theorem follows directly from theorem 1 : its truth 
becomes apparent on drawing a figure. 

Theorem 1' . If fix) is real, single-valued, and continuous through- 

fun like ) 
like \ s iy ns > 

then between a and b f(x) is zero an \ > number of times: the 

J K J ( even ) 

even number may of course be zero. 



§51 



SOME GENERAL PROPERTIES OF FUNCTIONS 



65 



The converse of this theorem is not universally true. That is, 
iff(x) = within an interval, we cannot assert that f(x) has both 

Y 
Y 





+ and — signs within that interval. In other words, if f(xi) = 0, 
we cannot assert that f(x) changes sign at Xi = 0, or that the 
graph crosses OX at X\. For plainly the graph may have one of 
the forms shown in the figures. To ascertain in such a case 
whether f{x) does actually change sign, we have only to deter- 
mine the signs of f(x\— h) and f(xi+ h) for a sufficiently small 
value of h. 

51. Roots of Polynomials. A root of a polynomial is any value 
of the argument which makes the polynomial zero. The roots 
may be real or complex. An equivalent geometrical definition 
for a real root is the following: 

A real root of a polynomial is the abscissa of a point in which 
the graph of the polynomial cuts the rr-axis. 

An algebraic equation is a polynomial put equal to zero, and a 
root of such an equation is any value of the argument which satis- 
fies it. Plainly, a root of an equation is the same thing as a root 
of the polynomial which constitutes the first member of the 
equation. 



Examples. 

1. 1 and 2 are roots of the polynomial x 2 
stituted for x they make the polynomial 
zero. Also, 1 and 2 are roots of the alge- 
braic equation 

x 2 - 3x+2= 0. 

The graph of the polynomial is shown in 
the figure. 



3 x + 2, because when sub- 




66 



DIFFERENTIAL CALCULUS 



§51 



2. The roots of the polynomial x 2 — 3x+3 are 



3±V-3 



because 



\F 




when substituted for x these values make the polynomial zero. They are 

also the roots of the equation x 2 — 3 z + 3 = 0. 
Since the roots are complex, they have for us 
no geometrical interpretation, except that 
they show that the graph does not cut the 
x-axis. The graph is shown in the figure. 

To find the real roots of a polynomial 
■ is thus seen to be equivalent to finding 
the points where its graph cuts the 
z-axis, and this is done by aid of a theorem V . How this theorem 
is applied to the solution of the problem is shown in the following 

example. 

Example. Let us find a real root of the polynomial 
P(x) = x* + x i -7. 

Solution. We find by trial two values of x that will give P{x) contrary 
signs. We find thus that P(l) = -5 and P(2) = 5. Hence, by theorem 
T, at least one root lies between 1 and 2 
(the polynomial is obviously single-valued, 
and no polynomial has a discontinuity). 
That is, the graph cuts OX between the 
point (1, - 5) and (2, 5). We next substitute 
1.5 for x and find that P(1.5) = -1.375, and 
hence by theorem 1' a root lies between 1.5 
and 2. 

We continue to substitute values for x as 
follows : 

P(1.6) = -0.344; hence the root lies between 1.6 and 2 
P(1.7) = 0.803; hence the root lies between 1.6 and 1.7 
P(1.65) = 0.215; hence the root lies between 1.6 and 1.65 
P(1.63) = -0.013; hence the root lies between 1.63 and 1.65 
P(1.64) = 0.1; hence the root lies between 1.63 and 1.64 

We have now found one root correct to the second place of decimals, 
viz., 1.63. 

By continuing the process we can find as many decimal places of the 

root as desired.* 

* In textbooks of algebra may be found a very expeditious method of making 
these substitutions, — a method known as Horner's Method for Approximating 
to the Roots of Equations. Horner's Method applies only to polynomials. 




by 
theorem 1'. 



52 



SOME GENERAL PROPERTIES OF FUNCTIONS 



67 



52. Zeros of Functions. A value of the argument which makes 
a function zero is termed a zero of the function. The terms root 
and zero are exactly equivalent, except that the term root is usu- 
ally confined to polynomials, while the term zero is applicable to 
any function. For example, 1 is a root of the polynomial x 2 — 1 , 

and is also a zero of it, while on the other hand - is a zero of cos x, 

but is not usually called a root of it. Theorems 1 and 1' apply 
of course not only to polynomials, but to all real, single-valued, 
continuous functions, and consequently the method of the preced- 
ing article may be applied to finding the zeros of functions that are 
not polynomials. 

Example. Let us find the zeros of f(x) = x — 2 sin x. 

To find the zeros of this function is equivalent to finding values of x 
that will make x = 2 sin x, and this is equivalent to finding the intersec- 
tions of the right line y = x with 
the curve y = 2 sin x. A good 
idea of the location of the zeros 
of x — 2 sin x may therefore be 
obtained by drawing the right 
line y = x and making a rough 
drawing of the curve y = 2 sin x. 
The loci plainly intersect at the 
origin. Moreover, with only a 
moderate amount of care in 
drawing the curve, it becomes apparent that the loci have only two 

other real intersections, and that one of them lies between - and v 

and the other between — - and — w. We can of course prove that 

such is the case by substituting these values for x in f(x) and noting 

that /(^j and /(tt) have contrary signs, as do also /( — |) and/(— v). 

Further, f(x) has no other positive zero, for 2 sin x cannot be greater than 
2, and hence any value of x greater than 2, such as tt, will make/(x) posi- 
tive. Similarly, it can be shown that there is no zero less than — x. Of 
course it was not necessary to draw the loci, but they enable us to judge 
what values to substitute for x in the more accurate determination of the 




X 


sin x 


/(*) = 

x — 2 sin x 


100° = 1.7453 


.9848 


- .2243 


110°= 1.9199 


.9397 


+ .0405 


105°= 1.8326 


.9659 


- .0992 


108°= 1.8849 


.9511 


- .0173 


108° 30'= 1.8936 


.9483 


- .0030 


108° 40'= 1.8965 


.9474 


+ .0017 


108° 37'= 1.8956 


.9477 


+ .0002 


108° 36'= 1.8953 


.9478 


- .0003 


Hence a zero of x - 


- 2 sin x is 108° 3( 



68 DIFFERENTIAL CALCULUS §§53-54 

Using now a table of natural sines, we form a table of values of x and f(x), 
and apply theorem 1'. 

Inference by aid of 
theorem 1'. 



zero between 100° and 110° 
zero between 105° and 110° 
zero between 108° and 110° 
zero between 108° 30' and 110° 
zero between 108° 30' and 108° 40' 
zero between 108° 30' and 108° 37' 
zero between 108° 36' and 108° 37' 



The other zero is of course — 108° 36'. 

53. Exercises. 

1. Find the positive root of z 3 — 6 x — 13. 

2. Find a positive root of x 4 - 12 x 2 + 12 x - 3. 

3. Find a root of 2 x z — 3 x 2 — 5 correct to three places of decimals. 

4. Find a zero of 3 x — tan x. 

5. Find a zero of x 2 — sin x. 

6. Find a zero of 2 X + x correct to three places of decimals. 

7. Find two zeros of x sin x— 1 correct to within 1'. 

8. Find all the zeros of x — sin x and x + sin x. 

9. Find a zero of x + logio x correct to three places of decimals. 

10. Consider the quadratic polynomial ax 2 + bx + c. Show that, when 
b 2 — 4 ac = 0, the polynomial has the same sign as c for all 
values of x. 

54. Increasing and Decreasing Functions.* 

Definitions. If, as x increases from a to b, f(x) always j l ™ creases I ? 

( an increasing ) . 

fix) is defined to oe{ 7 . } function in the interval atob. 

{ a decreasing ) 

-r,- { 1 ) ■ , • i • i • { increasing ) . 

Fig. < > is the graph of a function which is < , . > in 

the interval a to b. 

* Unless the contrary is expressly stated, the function is understood to be, 
real, single-valued, and continuous. 



§54 



SOME GENERAL PROPERTIES OF FUNCTIONS 



69 



Let P and Q be any two points of the arc AB, and let x, fix) 
be the coordinates of P. Then the coordinates of Q may be 
written x + Ax, f(x) + A/(x), where Ax is + or - according as 




Y 


^4 


Q 


p 


Q 







\w b X 




a 


X 


jBA 



Fig. 1. 



Fig. 2. 



Q is at the right or left of P. Then from our definitions and from 
the figures it follows that 

like ) 

unlike ) 

A/(x) 



( increasing ) 
If/(x)be j decreasin g5 along AB, Ax and A/(z) have 



Ax 



signs wherever P and Q may lie in AB, and that therefore 

is ] [ everywhere along the arc AB, that is, everywhere in the 

interval a to b. 



Now suppose Q to approach P as its limiting position, and 



A/(x) 
Ax 

to approach f'(x) as its limit. When a variable in approaching 
its limit maintains always the same sign, its limit either has that 
sign or is zero. Now none of the functions that we shall deal with 
have derivatives which are zero at an infinite number of points of 
a finite arc, except those functions which are constants, and which 
because of their triviality may be excluded from the present dis- 
cussion. We have, then, 



Theorem 2. If f{x) is 



increasing i 



along an arc, f(x) is 



+ 



decreasing ) 

at every point of that arc, with the possible exception of a finite number 
of points at which it may be zero. 

The geometric statement of the theorem is as follows: At every 



70 



DIFFERENTIAL CALCULUS 



§55 



7 • 7 *, n • \ increasing ) 
point of an arc, along which f(x) is < , . > , the tangent makes 

C acute ) 
with OX an < , > angle* except at such points as C (finite in 

number) , where the tangent is parallel to OX. 





Y 



/ \ 


\s^Cf 










\ X 


/ t 


I 




b a 


) \ 



More important for our purposes is the converse 

' + 



Theorem 3. If f(x) is 
C increasing ) 



at every point of an arc, fix) is 



, along that arc. 
( decreasing ) 

!QppT , pQC'IT"»p" i 
> in any part of the arc, fix) 
increasing ) J ' J v ' 



would, by theorem 2, be \ > or possibly at that part of the 
arc, and that is contrary to the hypothesis. 



Corollary 1. If f"{x) is 



\ + - 



at every point of an arc, f(x) is 



( increasing ) _ _ , 

<■ , > along that arc. 

I decreasing ) 

And, more generally, 

Corollary 2. If f n) (x) is ] __ > at every point of an arc, f n ~ l) (x) 

. I increasing ) 

is { -, } along that arc. 

( decreasing ) 

55. Convexity and Concavity. An arc of a curve y = f(x) is 

said to be \ [ when viewed from below (that is, when viewed 

concave \ 



At exceptional points it may be a right angle. 



55 



SOME GENERAL PROPERTIES OF FUNCTIONS 



71 



by an observer stationed at the foot of the page) when it has such 
shape as the arc AB, BC, or AC of Fig. j > . Let a point P 
traverse one of these arcs from left to right, carrying with it a tan- 




Y 






' V 


\ 






vi 


p^ 


\ 




/ 


A 


£ 




T| 


K 


/ 












X 




a 


b 


o 



Fig. 1. 



Fig. 2. 



gent to the curve. From the figures it can be seen that, if the arc 
convex 



be < > , as this tangent rolls along the curve, the angle 

( concave ) 

i • i • i • , ^r • ii \ increases ) , 

which it makes with OX continually < , > , and that 

( decreases ) 

.- . x . (an increasing ).,.,„ ^ 

therefore f (x) is < , > function along the arc. From 

f a decreasing ) 

the figures it can be seen that the converse of this is also true, viz., 

increasing ) . ( convex ) 

, . \ throughout an arc, that arc ^s < > 

decreasing ) ( concave ) 

when viewed from below. 

Applying corollary 1 of theorem 3, Art. 54, we have 

Theorem 4- Iff"(%) is ] [at every point of an arc, that arc 



If f'(x) is 



IS 



convex 
concave 



j when viewed from below. 



(convex ) 

< > arcs of a curve, y = fix). 

( concave ) ' u Jy n 



Hence, to determine the 

" concave 

we have only to determine the intervals within which fix) or y' 



is 



72 



DIFFERENTIAL CALCULUS 



§56 




56. The Flex.* A flex is a point at whicn a curve changes from 
convex to concave or from concave to convex. 
The curve in the figure has flexes at A, B, and C. Since the 

curve bends in opposite direc- 
tions on opposite sides of the 
flex, it is geometrically evident 
that the flex-tangent crosses the 
curve at the flex. 

There are cases in which the 
curve is convex on one side of a 
point and concave on the other 
side, and yet the point is not 
considered to be a' flex. For example, the tangent curve, y = tan x, 

changes from convex to concave at the points x = ± -, ± -^-, . . . 

(see figure in Art. 8), but since the ordinates of these points are 
infinite, that is, since tan x becomes discontinuous at these points, 
they are not classed as flexes. 

Again, the curve in the accompanying figure changes from con- 
vex to concave at P, and, moreover, has no discontinuity there. 
Yet P is not regarded as a flex. The peculiarity is that /'(a;) has 
at P a discontinuity like the discon- 
tinuity at D in the figure of Art. 49. 
Obviously the point P differs funda- 
mentally from the points A, B, C, above. 
Gn the other hand, f'(x) may have one 
kind of discontinuity at a flex; it may be 
infinite there, and then the only peculi- 
arity about the point is that the flex- 
tangent is perpendicular to the x-axis. Such is the case at B above. 

We must, therefore, modify our definition of a flex by saying 
that a point at which the curve y = f(x) changes from convex to 

* In most textbooks on the Calculus a flex is called a point of inflection. 
Professor Frank Morley first suggested the use of the short and convenient 
terms flex and flex-tangent, in place of the time-honored and cumbersome 
expressions "point of inflection" and "inflectional-tangent." 




§57 SOME GENERAL PROPERTIES OF FUNCTIONS 73 

concave, or vice versa, is not to be classed as a flex if fix) is 
discontinuous at the point, or if f'(x) has there a discontinuity 
other than an infinity, in other words, if the point is an angular 
point. 

From our definition of flex, and from theorem 4, it follows 
immediately that/"(z) changes sign at a flex, and that, conversely, 
wherever fix) changes sign is a flex (with the exceptions already 
pointed out). Hence, 

To determine the flexes of a curve y = fix), we have only to find the 
points at which fix), or y" , changes sign. 

Now fix) can change sign by passing through Oor oo, and in 
most of the elementary functions the second derivative can change 
sign in no other way. Hence the abscissas of the flexes of the 
curve y = f(x) are to be sought among the zeros and infinities * of 
fix), that is, by solving the equations 

f"{x) = and f'{x) = oo. 

Suppose £i to be a value of x so found. Then, by substituting 
in f{x) values of x a little less and a little greater than Xi (this 
substitution can often be made by inspection), we ascertain 
whether f" ix) does actually change sign at x. If it does so, fixi) 
must be calculated, and then the point fe, fixi)) is a flex of the curve 

y=f(x). 

57. Examples of Finding Flexes and Convex and Concave 

Arcs. 

i 
Example 1. The curve y % = x 3 (4 - x). 

Solving for y and differentiating twice, we have 

(a \i r 4 3 — z „ 4 x — 6 
y=xi±-xy y =-- -,y" = -- -. 

a (4 - x) a y (4 - x) 3 

It is obvious that y" changes sign at x = 4 and at x = 6 (the infinity and 



* An "infinity" of a function is a value of the argument that makes the 
icti< 
logs. 



function infinite. Thus x = 2 is an infinity of = ; x = is an infinity of 

x - - — 






74 



DIFFERENTIAL CALCULUS 



57 



zero of y") and at no other point. The ordinates of these points are 
and —6 \/2= —7.56 respectively. Consequently the points (4, 0) and 
(6, —7.56) are flexes, and they are the only ones. In order to know the 
shape of the curve at the flexes, we must know the directions of the flex- 
tangents. When x = 4, y' = oo , and when x = 6, y' = -2 \^2 = -2.52. 
Through the respective flexes we now draw lines with these slopes, and 
these are the flex-tangents. They mark the direction of the curve through 
the flexes. Furthermore, 

when x < 4, y" is — , and the arc is concave, by theorem 4; 

when 4 < x < 6, y" is +, and the arc is convex, by theorem 4; 
when x > 6, y" is — , and the arc is concave, by theorem 4. 



These data enable us to form a good 
idea of the general shape of the curve. 
It is obvious that the curve passes 
through the origin, and is readily found 
to have there the slope y r = \^4= 1.587. 
By plotting a few additional points, 
such as the points whose abscissas are 
— 1, 2, 3, 5, 7, the curve may be drawn 
with considerable accuracy. 

Example 2. The curve y = — • 

x 3 

Differentiating twice, we have 
3_ „ = 12 




y" has no zero, but it has an infinity, x = 0, and it is obvious that y' 

changes sign at x = and nowhere else.* But 

when x=0,y= oo , and the "infinite point " of 

the y-axis is a discontinuity and not properly 

a flex. When x < 0, y" is — , and when x > 0, y" 

is +, and consequently, by theorem 4, the curve 

is concave at the left of the origin and convex 

at the right of it. It exists only in the first and 

third quadrants. By plotting a few points it 

can be drawn with considerable accuracy. It 

resembles the equilateral hyperbola y = - • 




* y" also changes sign at x 
excluded from the discussion. 



oo, but points whose abscissas are oo are 



58 



SOME GENERAL PROPERTIES OF FUNCTIONS 



75 



Example 3. The semicubical parabola, y 2 = x 3 . 
Solving for y and differentiating twice, we have 



y=±x*,y 



, 3 i „ ,31 

±-x2,y' =±- — 

2 4r 



The curve consists of two branches symmetrical as to the z-axis. When 
x < 0, y is imaginary, and therefore the curve is real only in the first 
and fourth quadrants. Considering each 
branch separately, we see that y" has 
no zero, and, although it has an infinity 
at x = 0, it cannot change sign at that 
point, and hence there are no flexes. 

Furthermore, when y is J "•" j , y" is > 



and hence the \q^ branch is every- 



where 



The two branches 




convex 

concave 

meet in the origin, and this is a peculiar point of the curve. For, when 
x = 0, y' = for each branch, showing that the z-axis is tangent to each 
branch at the origin. Since the curve does not extend to the left of the 
origin, it must have at that point the shape shown in the figure. Such 
a point on a curve is termed a cusp. 

58. Exercises. 

Determine the flexes and the convex and concave arcs of the following 
curves. Draw the curves, plotting additional points where necessary. 
In every case draw the flex-tangents. 

1. y=l+(x-Sy. 7 . 

2. |/ 3 =2x 2 . 8 

3. y=xKx-?>). ' 

4. y=x*(2-x). y ' 

5. (y-iy = (x + 2)*. 10 - 



i 
as 



X3±2/£ 

y = x — sin x. 

y = tan x ; y = cot x. 



_ r ±2n 



(n an integer). 



6. x z ± y z = a 3 . 



11. y=x M2n+1) {n an integer). 



CHAPTER VII 
MAXIMA AND MINIMA 

59. Definitions. That value of a function at which it changes 

{increasing) (decreasing) . 77 ( maximum value ) 
from { 7 . > to {. . >is called a < . . , > of the 

J ( decreasing ) (increasing ) ( minimum value ) 

. . ( maximum ) ' ■ 

function, or simply a{ . . } of the function. 

( minimum ) 

Equivalent geometrical definitions are the following: 

• • { ascends ) 

When a point in traversing a curve first < , , Ho a fixed point 

_ _ ( descends ),„,.. 
0/ me cwrye, ana men < > , mai jtzea poin£ is termed a point 

( ascends ) 

( maximum ordinate ) i maximum point ) 

of I . . , . . >, or simply a< . . > • 

f minimum ordinate ) ( minimum point ) 




_ _ _ . . ( Af, M' ) ( maximum points ) 

In the figure the points < , \ are < . . . , > . 

I m, m ) ( minimum points ) 

From the definitions and from the figure, it is obvious that a 

(maximum ordinate) . , , , . „ , (greater) , _ .. 

< . . ,. , > is (algebraically) < , > than the ordi- 

f minimum ordinate ) ( less ) 

nates of all the points in a sufficiently small interval extending 

76 



§60 MAXIMA AND MINIMA 77 

on each side of the point in question. A maximum is not neces- 
sarily the greatest value, nor a minimum the least value that a 
function can have. Indeed it is evident from the figure that one 
and the same function may have several maxima and several min- 
ima, and that some of the minima may be actually greater than 
some of the maxima. 

60. Maxima and Minima Determined by Means of the First 
Derivative. The definitions of maxima and minima, taken in 
conjunction with theorems 2 and 3 of Art. 54, give rise to 

Theorem 5. Wherever f{x) changes from \ [ t° \ [ is a 

( maximum ) . 

{ . . > of fix), and conversely. 

( minimum ) 

Hence, to find the values of x that will make f(x) a maximum or 

a minimum, we must find the values of x at which f(x) changes 

sign. Now for the simpler functions fix) can change sign only 

by passing through or go . Hence we must find the zeros and 

infinities oif(x), that is, we must solve the equations 

f(x) = and f(x) = oo. 

Having found a value of x that satisfies one of these equations, 
say x h we next ascertain, by calculation or inspection, whether 
f(x) changes from + to — or from — to + at Xi. Lastly, we cal- 
culate f(xi), the maximum or minimum value of f(x). 

These results may also be arrived at by an inspection of the 
figure of the preceding article. It is obvious that at the maximum 
points M and M' ', and at the minimum points m and m' , the tan- 
gents are either parallel to OX (f(x) = 0) or perpendicular to 
OX (f(x) =oo). Moreover, in the case of a maximum point, 
whether of the form M or M' ', the slope, f(x), is + at the left 
and — at the right, while in the case of the minimum points, m 
and m',f(x) is — at the left and + at the right. Thus theorem 5 
is verified. If f(x) has the same sign on both sides of the point 
at which it is or oo , that point is a flex with flex-tangents parallel 
or perpendicular to OX. Such are the points 7 2 and I±. 



78 



DIFFERENTIAL CALCULUS 



61 



It is possible that f{x) should change sign by a spring instead 
of by passing through or oo. See Art. 50 (b). The point may 

still be a true maximum or 
minimum point, as in the ac- 
companying figure. Curves 
having such discontinuities 
will not be met with in this 
book. 

Example 1. Let us find the 
maximum and minimum values 

of f(x) = 2x 3 -9x 2 +12x. 

Solution. f(x) = 6a: 2 - 18 x+ 12 = Q(x -l)(x - 2). 

The roots of f(x) = are 1 and 2. 

When x< 1, f(x) is + ; when 1 < x < 2, f(x) is - ; when x > 2,f(x) is" +. 
Therefore by theorem 5,/(l) = 5 is a maximum, and /(2) = 4 is a minimum. 
Example 2. Let us find the maximum and minimum values of 

/(:r.)=(z 2 -l)i 
4 x 




Solution. 



fix) = 



3(^_i)i 

f(x) has a zero at x = 0, and infinities at x = ± 1. 
Taking these in the order of their magnitude, — 1, 0, 1, we see that 
when x<— 1, f(x) is — ; 

when — 1 < x < 0, f(x) is +; then by theorem 5, f(x) is min. at x = — 1; 
when < x < 1, f(x) is — ; then by theorem 5, f(x) is max. at x = 0; 
when x > 1, /'(#) is +; then by theorem 5, f(x) is min. at x = 1. 

Therefore, the maximum value of /(x) is /(0) = 1, and the minimum 
values are/( — 1) = and/(l) = 0. 

61. Exercises. 

Find the maximum and minimum values of the following func- 
tions : 



1. /(£) = £ 3 -3z+6. 

3. y= z 3 +3x 2 +6z+2. 

4. y = sin z. 
2x 



u = 



1 



5. y = 



1 + z 2 



* 2 - 1 

(x-a})(x-V) 
x 
x+1 
z 2 -4z+l* 



§62 



MAXIMA AND MINIMA 



79 



9. Show that x(x 2 — 5)* has one maximum and one minimum. 
10. Show that x(x 2 - 7) 3 has two maxima and two minima. 

62. Employment of the Second Derivative in Determining 
Maxima and Minima. Let Xi be a zero oif'{x) = 0. If }"{x x ) is 
decreasing 



, , fix) is 17 .~"° \ , by theorem 3, corollary 1, of Art. 54, 

( + ) I increasing ) 

and, since f'(x) passes through at x h it must have been J _ [ 
just before x reached the value x h and < , f just afterwards. 



Consequently, by theorem 5, f(xi) is 
then, 



maximum 
minimum 



We have, 



Theorem 6. If f'(xi) = 0, and if /"(zi) is ] [ , f'(x) changes 



maximum 
minimum 
The truth of this theorem becomes more obvious when we attend 



from \ _ [ to < > at x h and f(x{) is 

te truth of this theorem becorj 
to its geometrical signification. To say that f"(xi) is 
that f(xi) = 0, is t( 
(a:i,/(^i))is a point on a 



+ 



and 



that f(xi) = 0, is to say that 
concave 
convex 

arc at which the tangent is parallel 
to OX. And it is evident from 



the figure that such a point 






m 




. maximum 
must be a < . . > point, 

minimum l 



Example. Let us determine the maxima and minima of 
/(x)^3z 4 - 16z 3 +18z 2 +6. 



* It is here tacitly assumed that if f"(x) is -J , [ at x h it is also 



for 



values of x within a sufficiently small interval extending on each side of x x . 
It can be proved that this assumption is warranted, provided only that f"{x) 
is continuous at Xi. 



80 DIFFERENTIAL CALCULUS §63 

Solution. fix) = 12 x 3 - 48 x 2 + 36 x = 12 x(x - l)(x - 3) 

f"(x)= 12 (3 a: 2 -8 a: +3). 
The roots oif{x) are 0, 1, 3. Substituting these in f"(x), we have 

/"(0) = 36 > 0, and hence, by theorem 6, /(0) = 6 is min. 
/"(I) = -24 < 0, and hence, by theorem 6, /(l) = 11 is max. 
/"(3) = 72 > 0, and hence, by theorem 6, /(3) = -21 is min. 

Whether the method of this or of the preceding article is to be 
preferred in any particular case depends upon which it is easier 
to apply, whether it is easier to calculate the second derivative 
and then to determine its sign for x h or to determine the sign of 
the first derivative for values of x a little less and a little greater 
than Xi. 

63. Exercises. 

Find maximum and minimum values of the following functions: 

1. f(x) = 3x b — 25z 3 + 60 x. 5. u = sin x + cos x. 

2. z=30 4 -40 3 +120+7. 6. f(e) = sin 2 0+cos0. 
Q _ (x - a 2 ) {x - b 2 ) 7. S = tan - sin 0. 

o. U — n • 

x 2 o sin 

, 8. u=- ■ 

4 o_ r l + tan0 

r 2 +r+3' 9. f{u)= 2sinw+cos2w. 

10. If fix,) = 0, f"(xi) = 0, and / ,,/ fa) ^ 0, show that the graph of fix) 
has a flex at Xi. 

' 11. If f{x x ) = 0, fix,) = 0, fix,) = 0, and / IV (^) is j ~ j , show that 

,, x . ( maximum ) 
f(xi) is j . . 

( mimmum ) 

12. If the first five derivatives of f(x) are at x h and if / VI (xi) is j j , 

i ±i i. j-/ \ - { maximum ) 
show that f{xi) is \ . . J • 

(minimum ) 

13. Find maxima and minima of /(s) = 5 x 6 — 18 x 5 + 15 x 4 + 5. 

14. Find maxima and minima of f(x) = 3 x 8 — 4 a; 6 + 7. 

15. Determine the sign oif"{x) in the neighborhood of a maximum and 
of a minimum of f(x) at which f{x) = oo . 

16. How could f"{x) be used in the determination of flexes? 

17. Regarding y as argument and x as function, give a geometric 
meaning to x'(= D y x). Using x' and x" instead of y' and y", explain 
how convex and concave arcs would be defined. Show how, in such a 
case, to determine convex and concave arcs, flexes, and maximum and 



§64 MAXIMA AND MINIMA 81 

minimum points. Consider the case corresponding to the case in our 
general discussion where y' is oo at a maximum or minimum point. 

64. Curve Tracing. When the equation of a curve is not too 
involved, the foregoing methods enable us to locate the flexes, 
flex-tangents, convex and concave arcs, and the maximum and 
minimum points. And with these data in hand we are able to 
draw the curve with considerable accuracy. At least we know 
from the data the general shape of the curve, and further accuracy 
to any degree can of course be attained by plotting additional 
points. Any information that can be gotten by inspection of the 
equation or by simple calculation should be made use of. Thus, 
it is useful to note whether the curve is symmetrical as to the 
axes or as to the origin. In case there are intervals on the x-axis 
in which y is imaginary, this fact will have been brought to light 
by the examination for flexes and for maximum and minimum 
points, and these intervals should be marked. It is usually better 
to determine convex and concave arcs and flexes before examin- 
ing the equation for maximum and minimum points. 

Examples. 

1. 100 y = x b - 30 x z + 85 x = x(x*- 30 x 2 + 85). 

Solution. 

(a) Since odd powers of both variables occur, the curve is not symmetri- 
cal with respect to either axis. The equation is unchanged when we write 
—x for x and — y for y, and the curve is therefore symmetrical with 
respect to the origin. 

(6) Differentiating the given equation, we have 

20 y' = z 4 - 18 x 2 + 17 = (x 2 - l){x 2 - 17), and 5 y" = x(x 2 - 9). 

The roots of y" = are 0, ±3, and it is evident that y" does change sign 
when x passes through these values. Calculating y and y' for these values 
of x, we have the following flexes and slopes of flex-tangents: 

Flexes, (0,0), (3,-3.12), (-3, 3.12). 

Slopes, 0.85, -3.2, -3.2. 

When x< — 3, y" is — , and the arc is concave. 

When — 3 < x < 0, y" is +, and the arc is convex. 
When < x < 3, y" is — , and the arc is concave. 
When x > 3, y" is +, and the arc is convex. 



82 



DIFFERENTIAL CALCULUS 



64 




These intervals should now be marked and the flexes and flex-tangents 
drawn as in the figure. 

(c) The maximum and minimum points are next to be determined. 

When y' = 0, x = ± 1 or ± Vl7 = 
±4.12+, and it is evident that y' 
actually changes sign for these values 
of x. Moreover, when x = ±1, y" is 
=F, and when 3 = ±Vl7, y" is ±. 
Hence, 1 and — Vl7 are the abscissas 
of maximum points, and —1 and Vl7 
of minimum points. Calculating y for 
these values of x, we have 

maximum points (1, 0.56), (—4.12, 5.6), 
minimum points ( — 1,-0.56), (4.12,-5.6). 

These points are now plotted. 

(c) Solving 3(3 4 - 30 3 2 + 85) = 0, we have 0, ±5.18, ±1.78 as the 
abscissas of the points where the curve cuts OX, and the slopes at these 
points are respectively 0.85, 12.7, —1.5. After calculating the coordi- 
nates of a few intermediate points the curve may be readily drawn. 

2. y 3 =x i (2x-5)\ 

Solution. 

(a) The curve is symmetrical with respect neither to the axes nor to 
the origin. 

(6) Writing the equation in the form 

7/ = £3(2 x — 5) 

and differentiating it, we get 

,' 10 3-1 , „ 10 2Z + 1 

y = —-^r-> and y =— — — ■ 

3 33 9 33 



y" changes sign when x= —\ but not when 
3=0. In other words, by solving y" = we 
get the abscissa of a flex, but not by solving 
y"= oo. The flex is (-§,-3^2) and the 
slope of the flex-tangent is 5 \^2. It is readily 
found that the arc at the left of the flex is con- 
cave, and that at the right convex. 

(c) The solution of y' = and y' = oo are respectively 3 = 1 and 3=0, 
and plainly y' does change sign at these values of x. Moreover, when 
3=1, y" >0, and the point (1,-3) is a minimum point. When x= 0, 




§§ 65-66 



MAXIMA AND MINIMA 



83 



y" = oo , and here the second derivative gives us no information. How- 
ever, from the form of y' we readily see that as x passes through 0, y' 
changes from + to — , and therefore the origin is a maximum point. Since 
the tangent at this point is perpendicular to OX (is in fact OY), the origin 
must be a cusp. By plotting a number of additional points the curve 
may be accurately drawn. 

65. Exercises. Determine convex and concave arcs, flexes, 
flex-tangent, and maximum and minimum points, and draw the 
curves given by the following equations: 



1. 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 

66. 



14. 3 



15 



Sy=x(9-x 2 ). 
8 y = x 4 — 4 x 3 . 
4:y=8x 2 -x\ 

y={x z +l)(x-2). 
20y = x*-80x. 
5y=3x 5 -10x z . 

y Z = X 1 — 1. 

y 2 = x z — 1. 
y z = x A - 1. 
100y = x(25-x 2 ) 2 . 
y z = x 2 (x + 5) 3 . 
5$y=x(7-x 2 ) z . 
y z =x 2 {x 2 -4) z . 

Problems in Maxima and Minima. 



l-3z 



2x 



1 + x 2 

16. b 2 x 2 =b a 2 y 2 = a 2 b 2 . 

17. y= sin x; y = cosz; 





y= secx; y = esc a:. 


18. 


The astroid x^ -f- ya = 


19. 


2 2 2 
£3 _ y 3 = a3 . 


20. 


y 2 = a 2 sin x. 


21. 


y z = a 3 sin x. 


22. 


y 2 = a 2 sin 3 x. 


23. 


y z = a 2 sin 2 x. 


24. 


y z = a z tan x. 


25. 


y = 2 sin x + cos 2 z. 



as. 



(a) A square of tin whose side is a has a small square cut from each 
corner, and then the sides are bent up to 
form a box. Determine the side of the 
small squares so that the box shall have 
the maximum cubical contents. 

Solution. Let x be the side of the small 
squares cut away, and let v be the cubical 
contents of the box. Then 

v = x(a — 2 x) 2 = a 2 x — 4 ax 2 + 4 z 3 , 

and we are to determine x so that v shall 
be maximum. Differentiating, 

y'= D X V= O 2 - 8CKE+12Z 2 , 

v"=D x 2 v=-8a+24:X. 



X 
X 




X 
X 




a-2x 




X 
X 




X 
X 



84 



DIFFERENTIAL CALCULUS 



§67 



The solution of v' = is x = - or - • 

o I 

9 

When x = -, v" is — , v = — a 3 , the maximum contents. 
6 '27 

When a; = - , v" is +, v = 0, the minimum contents. 

(b) Find the right circular cylinder of greatest volume that can be 
inscribed in a given right circular cone. 
Solution. Let r be the radius of the base of the cone, and h its altitude, 

and let x and y be the radius and alti- 
tude of the inscribed cylinder. Let v 
be the volume of this cylinder. 

Then v = irx 2 y. From similar triangles 

— - — = -, whence y= - (r—x) and 
r - x r r 

v = — [r — x)x 2 . 
r 

We are to determine x so that v shall be 

a maximum. If we place u = (r — x)x 2 , 

it is evident that, since — is a constant, 
r 

v is a maximum or minimum whenever 

u is. Therefore, to avoid writing the constant factor — , we shall deter- 

r 

mine the value of x which will render u a maximum. 

u=rx 2 -x z ; u' = 2rx-3x 2 ; u" = 2r-6x. 
When u' = 0, x = f r or 0. 

When x = f r, w" is — , v = A ^h, the maximum volume. 
When x = 0, w" is +, y = 0, the minimum volume. 

67. Exercises. 

1. Separate the number a into two parts whose product shall be a 
maximum. 

2. Separate the number a into two parts the sum of whose cubes shall 
be a minimum. 

3. Show that among rectangles of given perimeter the square has the 
greatest area. 

4. Show that among rectangles of given area the square has the least 
perimeter. 

5. Show that among rhombi of given area the square has the shortest 
side. 




§87 MAXIMA AND MINIMA 85 

6. Show that among rhombi of given side the square has the greatest 
area. 

7. A circular sector of given perimeter, a, is to have a maximum area. 
Find radius and area of the sector. 

8. Find the right circular cylinder of greatest convex surface that can 
be inscribed in a right circular cone. 

9. Find the right circular cylinder of greatest total surface that can 
be inscribed in a right circular cone. 

10. Find the right circular cylinder (1) of greatest volume, (2) of 
greatest convex surface, that can be cut from a sphere of radius r. 

11. Find the right circular cone (1) of greatest volume, (2) of greatest 
convex surface, that can be cut from a sphere of radius r. 

12. Determine the relative dimensions of the right circular cylinder 
(1) of given volume and of minimum total surface, (2) of given total sur- 
face and of maximum volume. 

13. A cylindrical cup of given volume is to be made with the least 
amount of tin. What must be its relative dimensions? 

14. Let a be the slant height of a right circular cone. Find the other 
dimensions so that the volume shall be a maximum. 

15. A segment of the parabola y 2 = 6 z is cut off by a double ordinate 
x = 18. Find the dimensions of the rectangle having two of its vertices 
in the double ordinate and the other two in the curve, and (1) whose area 
is a maximum, (2) whose perimeter is a maximum. 

16. Let the figure of 15 be revolved about OX. The parabola generates 
a curved surface termed a paraboloid of revolution, and the rectangle 
generates a right circular cylinder. Determine the dimensions of the 
rectangle so that 

(1) the volume of the cylinder shall be maximum; 

(2) the convex surface of the cylinder shall be maximum. 

17. Let a rectangle be inscribed in the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 . De- 
termine the dimensions of this rectangle so that 

(1) the area shall be maximum; 

(2) the perimeter shall be maximum. 

18. Let the figure of 17 be revolved about bX. The ellipse generates 
a curved surface termed an ellipsoid of revolution, and the rectangle 
generates a cylinder. . Determine the dimensions of this cylinder so that 
(1) its volume, (2) its convex surface, shall be maximum. 

19. Determine the tangent to the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 , so that 

(1) the portion intercepted by the axes shall be of minimum length; 

(2) the area of the triangle made with the axes shall be minimum. 

20. A wall 8 feet high runs parallel to the side of a house and 7 feet from 



86 DIFFERENTIAL CALCULUS §67 

it. What is the length of the shortest ladder that will reach the house from 
the ground outside the wall, and how high up on the house will it reach? 

21. Two passages of widths a and b meet at right angles. What is the 
length of the longest rod that can be passed horizontally and without 
bending from one passage into the other? 

22. A circular piece of cloth 24 feet in diameter has a sector cut out of 
it, and the remaining sector is then made into a conical tent covering. 
What must be the angle of the sector cut out in order that the tent shall 
have the greatest capacity? 

23. Solve exercise 22 when 2 a is the diameter of the circular piece of 
cloth. 

24. At what distance from the side of a house must a man, whose eye is 
5 feet from the ground, station himself, in order that a window 7 feet high, 
whose sill is 14 feet from the ground, may subtend the greatest visual 
angle? 

25. At what height on the wall of a room must a source of light be 
placed so as to produce the greatest brightness at a point on the floor at 
a distance a from the wall? The brightness of an illuminated surface 
varies inversely as the square of the distance from the source of light, and 
directly as the cosine of the angle which the rays make with a normal 
(perpendicular) to the surface. 

26. Two spheres of radii r and R have their centers at a distance d 
apart. Where must a light be placed on the line joining their centers so 
as to illuminate the greatest surface possible? 

27. A platinum dish is to be made in the form of a spherical segment. 
What must be its relative dimensions in order that it may have a given 
capacity with the least amount of platinum? 

28. A wineglass is filled with water and a marble is carefully dropped 
into it. If the part of the glass above the stem is in the shape of an in- 
verted cone, the diameter of whose base is 2 a, whose altitude is b, and slant 
height s, show that the radius of the marble that will cause the greatest 

overflow of water is — -— • 

(s - a) (s + 2 a) 

29. A man in a boat a 'miles from A, the nearest point on a straight 
shore, wishes to reach, in the shortest time, B, a point on th^shore b miles 
from A. He can row r miles per hour and walk s miles per hour, (s > r). 

Show that he must land at a point whose distance from A is — == • 

Vs 2 - r 2 

Note that this result is independent of b, and explain its meaning when 

Vs 2 - r 2 



§67 MAXIMA AND MINIMA 87 

30. It is found that the value of the marble in a certain quarry in- 
creases as the square root of the depth below the surface. However, the 
cost of quarrying increases directly as the depth. If at a depth of 1 foot 
the marble sells at $1.80 per cubic foot, and the cost of taking it out 
is 15 cents per cubic foot, find from what depth the greatest profit will be 
secured, and find what that profit will be. 

31. If the total cost per hour of running a certain steamboat is propor- 
tional to the cube of the speed in miles per hour, what is the most economi- 
cal rate (the rate that will give the minimum cost per mile) to run her 
against a current flowing 6 miles per hour? 

32. The cost per hour of the fuel in running a certain steamboat is 
proportional to the cube of the speed, and at a speed of 12 miles per hour 
the cost of the fuel is $24 per hour. If other expenses amount to $162 
per hour, what is the most economical rate to run her in still water? 

33. Solve exercise 32 on the supposition that the boat is to run against 
a 6-mile current. 

34. Solve exercise 32 on the supposition that the boat is to run with 
a 6-mile current. 

35. Two trains A and B are running on two straight tracks which inter- 
sect at right angles at C. Their rates of speed are respectively 30 and 40 
miles per hour. At noon A is 100 miles from C and B 50 miles from that 
point. Find their time of nearest approach, and their respective positions 
at that time, 

First, when both trains are moving toward C; 

Second, when both trains are moving away from C; 

Third, when one is moving toward and the other away from C. 

36. A voltaic battery has an electromotive force of E volts and an 
internal resistance of r ohms. When the battery sends a steady current 
through an external circuit of R ohms resistance, w, the amount of work 
done each second in the external circuit, is proportional to 

E 2 R 

{r + R) 2 ' 

Give different values to R and show that the greatest efficiency is obtained 
(w a maximum) when R = r. 



CHAPTER VIII 

DIFFERENTIATION OF LOGARITHMIC AND EXPONEN- 
TIAL FUNCTIONS. DIFFERENTIALS 

68. To Determine lim [l-\ — ) . It might seem at first glance 
that we could write 

il»( 1+ ;DH + ^r = (1+0)CO=r = 1 - 

But, in the first place, we have no warrant for using co like an 
ordinary number, in any calculation, and, in the second place, it 
can be shown beyond question that the limit is quite different 
from 1. This we shall now prove. 

We first restrict a to be a positive integer. We may then expand 
the function by the binomial formula and get 

a(a - 1) 1 



('+i)'=-l3 



+ 



]2 a* 



a(a -l)(a- 2) 1 1 

"1 iq 3" T • • • -r~» 

[o a 6 a a 



or 






, (-S('-D('-9 , + , 

+ [4 + * ' •+««' 



When a > 1 the numerators in the second member of (1) are all 
positive, and therefore f 1 +-] > 2 for all positive integral values 
of a above 1, that is, for all values (above 1) that a takes in ap- 



(2) limfl- = ■;■-. !- ; 



§68 LOGARITHMIC AND EXPONENTIAL FUNCTIONS 89 

proaching its limit, oo . Consequently lim f 1 + -J is greater than 

2. Now as a = oo each numerator in the second member of (1) 
has the limit 1 and the last term has the limit 0. Hence 

a) ~ 2 + £ + 0r [4 + |5 

If now the restriction that a be a positive integer be removed, 
that is, if a be allowed to run through any series of positive values 
in approaching oo , it can be proved that the function has still the 
limit given in (2). We shall not give the proof of this, but shall 
accept (2) to be true without any restriction upon a. The nu- 
merical value of the second member of (2) can be calculated to 
any degree of approximation. Thus 

2 = 2.000000 
,4= .500000 

§ = or 3== - 166666 

4 = 4-4= .041666 
i| = i-^5= .008333 

4=4^6= .001388 

[6 [5 

i = i-7= .000198 

[7 [6 

| = I-8= .000024 

k = h 9= • 000002 

Sum = 2.718277 



90 DIFFERENTIAL CALCULUS §69 

which is correct to the fourth decimal place. This limit is denoted 
by e, and the number e, like the number t, is of great importance 
and of frequent occurrence in mathematics. In the following 
formula the value of e is given correct to the eighth decimal place. 

(A) lim ( 1 + -X = e = 2.71828183. 

If in this formula we write - = ft there results 

a 

(B) lim (1 + p)$ = lim (l + -)° = e. 

0=0 a=oc\ Oi) 

Again, 

(C) limfl+-) a = limr/l+- 1 \^ 

a= x \ a/ o^oo [ a 

_\ mf 
Putting m = — 1 in (C), we have 

, 1 



(D) lim(l--Y 



e 



e 



69. Differentiation of Logarithmic Functions. The reader is 
assumed to be acquainted with the elementary properties of 
logarithms. Thus he is supposed to know that if u = a y , y is 
defined to be the logarithm of u to base a, and that this definition 
is expressed by the equation y = log a u. In the common or 
Briggs system of logarithms, the system used in trigonometry, 
and in all arithmetical calculations, the base is 10. In the Calcu- 
lus, however, and in advanced mathematics generally, the system 
most often used is that whose base is e, and this system is termed 
the natural or Napierian system of logarithms. Why the system 
whose base is this extraordinary number e, a number whose nu- 
merical value can only be approximately determined, should be 
termed the natural system, cannot be fully explained in an ele- 
mentary course. One reason for it will appear when we come to 
differentiate logu. In the Calculus, when a logarithm is used 
without mention of the base, the base e is always understood, so 
that log u means log e u and not logio u. 



§69 LOGARITHMIC AND EXPONENTIAL FUNCTIONS 91 

We shall now prove the following formulae: 
XXI. Z> log a u = log a e ; 2> \og a <c = log„ e • — 

Proof. Dlog a u = lim — A , 

A*=0 Ax 

A log a U = log a (w + Aw) - \0gaU = logJ 1 + —J 

11 

Aw , A . Au\ Au * 

= ir l0 H 1 + ir] • . 

Aw ^ 

„ A log a w As, / 1 . Aw\ A " 

Hence -A^ = - l0ga ( 1+ FJ ' 

Aw _u 

Ax / Aw\ Aw 
and D loga w = lim lim loga [H • • (5, Art. 20). 

Ax=0 W Aj;=0 \ W / 



Aw 



Now 


,. Ax 
lim — = 

Ax=0 W 


Z>w 

— > 
w 




and, by yj, Art. 20, 










Ai/\ A « 




u 

Ai/\ A « 



lim logaf H ) = loga lim ( H J 

Ar=0 V U ) 3 Ax=0\ m / 

w 
Further, when Ax = 0, Aw = 0, and -r— = oo . Therefore, writing 

w 
-r— for a in formula (A), Art. 68, 

lim / 1 + M\ lim ( 1 + A#« 

=00 

Au 

u 

I _| =log a e. 

w / 

Therefore, finally, 

D loga w = loga e • Q. E. D. 

w 

* From m log a A = log a A m follows log a A = — log a A m . 



92 DIFFERENTIAL CALCULUS §70 

The second part of the formula is gotten by writing x for u in the 
first part. 

If in formulae XXI we write e for a and note that log e e = 1, 

there results 

Du , „ 1 

XXII. Dlogu = — and D logcc = -• 

U QC 

Du 

— > the derivative of logu, is termed the logarithmic derivative 
u 

ofu (see Art. 26, V). 

The formula for the derivative of a product (V, Art. 26) may be 
obtained by aid of XXII. Thus, from the equation 

log uv = log u + log v, 

we get by differentiation 

D log uv = D log u + D log v, 

VVTT Duv Du Dv 

and, by XXII, = 1 » 

uv u v 

whence Duv = vDu + uDv. 

Problem 1. Prove the formula for Du n by aid of XXII. 

u 
Problem 2. Prove the formula for D - by aid of XXII. 

v 

70. Differentiation of Exponential Functions. 

XXIII. Da u = a tl log a Du; Da x = a 00 log a. 

Proof. Differentiating the identity 

log a u = log a • u, 

Da u 

we have — — = log a - Du by XXII 

a 

whence Da u = a u log a • Du. Q. E. D. 

Writing x for u in this formula, we get the second part. 

If in XXIII we write e for a and remember that log e = 1, we have 

XXIV. De tt = e u Du and De x = e x . 

e x is the only function whose derivative is equal to the function. 



70 



LOGARITHMIC AND EXPONENTIAL FUNCTIONS 



93 



Examples. 

1. Dlog a (x 2 +x+l) = log a e 



£>(x 2 +x+l) _ 2x+l 



x 2 +x+l x 2 +x+l 

2x+l 



log a e. 



n t^i ( 9 . . i\ D(x 2 +X + 1) 

2. D log (x 2 + x + 1) = — V; — - 1 - — - 

x 2 + x + 1 x 2 + x + 1 

3. 2)0^+*+! = a*+*+i loga£(x 2 +x + 1) = (2x + l)a^+x+i i oga> 

4. De^+z+^e^+^D^+x+lH (2a;+l)e^+x+i. 

_ ~ , • o o D sin 3 x 2 6 x cos 3 a: 2 fi , Q „ 

5. D log sin 3 x 2 = . = — . n „ = 6 x cot 3 x 2 . 

sm 3 x 2 sin 3 x 2 

6. Z)e sin 3 * 2 = e sin 3 * 2 D sin 3 x 2 = 6 x cos 3 a; 2 • e sin 3 * 2 . 

In differentiating logarithmic functions there is often advantage 

in using the formulae 

u 
log uv = log u + log y; log - = log u — log v; log w n = w log u. 

The following examples illustrate this fact. 

Dmu _ mDu _ Du 
mu u ' 



7. D log rait = 



but better thus: 



8. Dlog(xsinx) 



D log mu = D(\og m + log u) = 

u 

D x sin x _ sin x + x cos x m 
z sin # x sin x ' 



or thus : 



D log (x sin x) = D(log x + log sin x) = - + - — 

^ sin a; 



smx+a; cos x 
x sin a: 



9. Dlog 



x-1 
x+ 1 



D 



x- 1 

x + 1 



x+1 



x-- 1 

x+ 1 



(x+1) 2 x-1 x 2 -l' 



but better thus : 



Dlog^— | = Dlog(x-l)-Z)log(x+l) = -^-- 1— = -^ 



x+1 



1 x+1 



10. DlogVx 2 
but better thus: 



^ = DVx 2 - 1 = x(x 2 - 1)~* 



V. 



Vx 2 - 1 



V 



Dlog Vx 2 - 1 = Dhog(x 2 - 1) = 1 -f*- = -/-, 

2 2 x 2 — 1 x 2 — 1 



11. Dlog-L _Z>/_|log*)— * 

Vi V 2 / 2 x 



94 DIFFERENTIAL CALCULUS §§71-72 

A reason for regarding e as the natural base of a system of loga- 
rithms is that in differentiating log u and e u our results are free 
from a constant factor, which is not the case in any other system. 

Thus D loga u = — loga e, while D log u = — ; 

u u 

and Da u = a u log aDu, while De" = e u Du. 

71. Exercises. Perform the following differentiations: 

1. D^+z*; D x e*+i*. 13. Dlog(w±V^T). 

2 . D x u loga m ; D x u log W. eX 2 _ j 

3. D x ya?;D x yey. 14 - ^g^y 

4. AW; ZW. 15. ^[log(^ + i) 2 -2,l. 

5. De«;D5-3*\ 16 z^-logtf 2 ). 

6. D log sin 0; D log esc 0. 

7. Z> log cose; D log tan 9. 17 - ^^ Wx+Vx + 2). 

8. De^; D2 V ". 18- D2[Vx-log(l+Vx)]. 

9. Da«*»» ; £e tan » 19> ^ (tan ^ + 2 l og cos 0). 
10. Due coau ; De u cosu. , 

tV^T 20. Z)log Va;+1 — . 

il. DlogVy V*+l+l 

1+VS 21. DUtan-^-log vV+l). 

12. Dlog— - — — • 

1-Vx 22. Z>z 3 (3logz- 1). 

Find the maximum and minimum values of the following 
functions : 



23. z 2 -logz 3 . 


28. 


xh x . 


24. z 2 log£. 


29. 


x 4 e x . 


25. x 3 log x. 


30. 


xe x . 


26. sin 6 - e sin e . 


31. 


x z e x \ 


27. zV. 


32. 


z 2 -log(z 2 +l) + 2. 


Draw the following curves: 






33. y = xe x . 


37. 


y = log sin x. 


34. y = log x. 

35. y = e. 


38. 


y=x—e x . 

i 


36. y=x— log Z. 


39. 


y = e x . 



72. The Compound-interest Law. The formulae of Art. 68 
can be applied to the solution of many problems. For example, 



§72 LOGARITHMIC AND EXPONENTIAL FUNCTIONS 95 

the amount of $1000 at 8 per cent interest for one year is $1000 
X (1 -f- 0.08). If the interest be compounded semiannually, the 

(0 08\ 2 
1 H — ~ ) ; if compounded three 

/ 0.08V 
times in the year the amount is $1000 f 1 -\ — — J ; and if com- 
pounded n times during the year, the amount at the end of the 

/ 0.08\ n 
year is $1000 f 1 -\ J Now by giving suitable values to n we 

find what the amount in one year will be if the interest is com- 
pounded weekly, daily, hourly, etc., and by conceiving n to become 
infinite we reach the idea of interest being compounded continu- 
ously. In that case the amount in one year is given by the 
formula 

A = lim$100o(l + — V 
Now by formula (C) of Art. 68, 






lim(l +**)"- 



Hence* 

A = $1000 e 008 = $1083.30. 

i 



y the same method of reasoning we arrive at the general formula 
or the amount in one year of a principal P compounded contin- 
uously at r per cent per annum, 



A = limP(l+~) n =Pe™. 
„=oo \ 100 n) 

The amount in t years is given by the formula 



A t = Pe m . 

This equation is what is termed the compound-interest law. A t 

kis function and t is argument and the equation is of the form 
y = Ce ax , 
where C and a are constants. A characteristic property of such a 

* Logio e = 0.4343. 



96 



DIFFERENTIAL CALCULUS 



§72 



function is that its derivative is proportional to the function itself. 



For 



D x y = Cae° 



ay. 




There are many phenomena which proceed according to the com- 
pound-interest law. Following is an example. 

Let R be a reservoir fitted with two pipes P x and P 2 , through 
which liquid can be admitted to and drawn from R. Let Pi and 
P 2 be of such relative sizes that the amount of liquid that flows 
in through Pi in a time t is exactly equal to the amount that flows 

out through P 2 , so that when both pipes 
are open the level of the liquid in R 
remains unchanged. Suppose R at the 
start to be filled with water, and sup- 
pose alcohol to be admitted through Pi 
while P 2 is left open. Suppose, further, 
that a stirrmg mechanism keeps the 
water and alcohol thoroughly mixed, 
so that the liquid that flows out through P 2 is completely homo- 
geneous. The level of the liquid in R will remain unchanged,* 
and the problem is, given the rate of flow through the pipes to 
determine the proportions of water and alcohol in R after a time t. 
Let v be the volume of water in R at the start. Let jw (/* a 
positive proper fraction) be the volume of liquid that flows through 
either pipe in a unit of time. Let both pipes be opened for a 
small interval of time, 6, and then closed. Suppose the stirring 
mechanism not in operation during the time 6, so that nothing 
but water flows out through P 2 . The amount of water that flows 
out is fjiv9 and the amount left in R is v(l — nd). Of course the 
amount of alcohol that flows in through Pi is fxvd. Now let the 
liquid be stirred until it is homogeneous, and then let both pipes 
be again opened for a time 6. Just as much liquid flows out 
during the second time interval as before, but only a part of it is 
water. And, just as the volume of water in R at the end of the 

* To simplify the problem, we leave out of account the densities of the 
liquids and assume that the volume of the mixture in R is equal to the sum of 
the volumes of the water and alcohol that it contains. 



§73 DIFFERENTIALS 97 

first time interval was 1 — fid times the volume at the start, 
so now at the end of the second time interval the volume of water 
in R is 1 — fid times the volume at the beginning of this interval, 
that is, is v(l — fid) 2 . Repeating this process a third time, it is 
found that the volume of water in R at the end of time 3 is 
v(l — fid) 3 . And, in general, after n repetitions of the process, 
the volume of water in J? is w(l — fid) n . Let t be the whole time, 

so that t = nd, and 6 = - , and let V t be the volume of water in R 
n 

at the end of time t. Then 



V t = v( 



1-^Y 
n) 



Now by letting 6 decrease indefinitely while n increases indefinitely 
we arrive at the idea of a continuous process, in which the flow 
through the pipes and the stirring of the mixture are simultaneous 
and continuous. In that case, by formula (C) of Art. 68, 



V t = \\mv(l -^Y 
n=oo \ nj 



— = ver**. 



The volume of alcohol in the mixture at the end of time t is of 
course v — ve-* 1 = v(l — e"^ 1 ). 

Problem 1. What is the amount of $2735 in 3 years if interest is com- 
pounded continuously at 6 per cent per annum? in 4 years? 

Ans. $3274.39. $3476.92. 
Problem 2. Suppose the reservoir, R, to contain 10 liters of water at the 
start, and suppose the rate of flow through each pipe to be 2 c.c. per second. 
What is the amount of water in R after 1 hour? after 2 hours? 

Ans. 4.867 liters. 2.36927 liters. 

73. Differentials. 

Definition. The product of the first derivative of a function by the 
increment of its argument is defined to be the differential of the func- 
tion. 

The differential symbol is d and the differential of f(x) is written 
df(x). Our definition may now be expressed as follows: 

(1) df(x)=f(x)Ax or Df(x)Ax. 



98 



DIFFERENTIAL CALCULUS 



§73 



If fix) is x itself, we have 

dx = x'Ax or DxAx, 
or, since x' = Dx = 1, 

dx = Ax, 

that is, the differential of the argument is identical with the increment 
of the argument. 

Writing dx for Ax in (1), we have 
(2) df(x)=f(x)dx or Df(x)dx* 

If y be the functional symbol instead of fix), we have 
(2 r ) dy = y' 'dx or Dy dx* 

Hence in the definition the word increment may be replaced by 
the word differential. 

There is a very simple geometric representation of df(x) or dy. 




&x=dx 



r 




T A 


/ 




A 


]df(x) 
= dy 




i> 










Ax=dx 


B 




//fix) 


=y 







Ap 


&x~dx 


X 



In the right triangle PBT, 

BT = PB tan 0. 
But PB = Ax = dx and tan 6 = f'(x) or y' . 

BT = f'{x)dx or y'dx, 
that is, BT = df(x) or dy. 

Hence, differential f{x) is the increment of the ordinate of the tangent 

to the graph of f(x) at the point (x, fix)), dx can be chosen at will. 

It is manifest, then, that the differential of the function is quite 

different from the increment of the function. 

* Because of these relations the derivative was formerly termed the differ- 
ential coefficient, a cumbersome designation that is rapidly falling into disuse. 



"§73 DIFFERENTIALS 99 

If formulae (2) and (2') be divided by dx, there results 

(3) ^sr-rM-AK*) and &mtfmDy. 

In the first members of these equations we have a new notation for 
the derivative. In any formula involving derivatives of functions 

of x such as Du, Dv, we may replace the latter by -=-, -=-, and 

then, clearing of fractions by multiplying by dx, we have the cor- 
responding differential formula. For example, from 



we get 



n u _ v Du — uDv 

V V 2 

,u du dv 

v dx dx . , u vdu — udv 

-i- = ■ and a- = - 9 , 

ax v l v v l 

the last being the differential formula corresponding to the given 

derivative formula. 

Again, from 

Dx n = nx n ~ l 

we get 

d(x n ) 
, = nx n ~ l and d{x n ) = nx n ~ l dx. 

The differential notation is very widely used. In fact, it is as 

du 
common to find the derivative of u expressed by -r- as by Du or 

D x u, and to find differential formulae as to find derivative formulae. 
For this reason we shall give here some of our formulae of differ- 
entiation expressed in the differential notation. The student can 
readily supply the omitted formulae. 

i. £-i. 

dx 

dr 

II. ^ = and dc =0. 

dx 

TXX dcu du , , , 

ill. —r- = c^r and dcu = cdu. 

dx dx 



100 DIFFERENTIAL CALCULUS §73 

TV d(± u zL v =b . . . ) _ du dv 
dx dx dx 



and d(± u ± v ± . . .) = =L du =L dv =L . . . . 

duv _ du 
dx dx 



T7 duv du , dv , 7 7,7 

V . -=— = v -; — f-Wr and cmy = vau -\- udv. 

dx dx dx 



VI. 

VII. 


— 7 — — nw- * -T- 
dx dx 

tU du dv 

d— v-. u-r 

v dx ax 

dx v 2 


ana 
also, 

and 


ayu") — nw* x au; 
d{x n ) = nx n ~ 1 dx. 

^u vdu — udv 
d- = ^ 

V v 2 


VIII. 


,1 dv 

v dx 

dx v 2 


and 


,1 dv 

V V 2 


IX. 


d sin u du 

— 7 — = cos u ~r 
dx dx 


and 


d sin u = cos u du. 


XV. 


du 
dsin~ 1 u dx 


and 


7 . , du 


dx Vl - u 2 


Vl - u 2 


XXI. 


du 

d\og a u _ , dx 

dx u 


and 


7 . , du 
d loga u = loga e — • 
u 


XXII. 


du 

d log u dx 

dx u 


and 


71 du 

d log u = — • 
u 


XXIII. 


da u , du 

Hx =aUlosa Tx 


and 


da u = a u log a du. 


XXIV. 


de u u du 
dx dx 


and 


de u = e u du. 



The equations of tangent and normal to the curve y = f(x) at 
the point (x 1} i/i) (see Art. 23) become, in differential notation, 

Tangent, y — y x = -p (x — x ± ) and (y— yi)dx x — (x — x 1 )dy l = 0. 
Normal, y — yi = — -j- (x — x x ) and (y — yi)dyi+ (x — x{)dxi = 0. 

dyi 

dxi 



§§74-75 DIFFERENTIALS 101 

74. Differentials of Higher Orders. 

Definition. The product of the second derivative of a function by 
the square of the increment (differential) of the argument is defined to 
be the second differential of the function. 

The symbol for the second differential of f(x) is d 2 f(x), so that 
our definition may be expressed thus (remembering that A# = dx) : 

d 2 f(x) =f"(x)dx 2 = D 2 f(x)dx 2 * 
and if y be the functional symbol, 

d 2 y = y" dx 2 = D 2 y dx 2 .* 
Dividing these equations by dx 2 , we have 

^M ./»(*) s DV(*) and g- 2," - Z% 

which is a new; notation for the second derivative. 

Definition. The product of the r th derivative of a function by the 
r th power of the increment (differential) of the argument is defined 
to be the r th differential of the function. 

The rth differential of f(x) and y are written d r f(x) and d r y, 
so that we have 

d r f(x) =f^(x)dx r = D r f(x)dx r 
and 

d r y = y( r ) dx r = D r ydx r . 

Dividing these equations by dx r , we have 

iM:^ mx) . D«/(x) and g - 2/W - Z> V , 

which are new notations for the r th derivative. 
Problem. Show that where x is the argument d r x = for r > 1. 

75. Remarks on the Differential Notation. It is evident that 
differentials might have been introduced much earlier in the 
course — in fact, at any stage after the definition of the derivative 
in Art. 22. The reader must not infer that they are introduced 

Lhere because any particular advantage is to accrue from their use 
in the following pages. The differential notation possesses few 
* dx 2 = (dx) 2 and more generally dx r = (dx) r . 



102 DIFFERENTIAL CALCULUS §75 

advantages over the notation that we have employed up to this 
point, and throws no new light upon any of the problems of 
mathematics. Indeed it has difficulties of its own and can pro- 
duce confusion of mind in the beginning student. But it is a 
part of our mathematical inheritance from the early days of the 
Calculus, and is still widely used, especially by writers on applied 
mathematics. Because, then, of their use in modern textbooks, 
and for this reason only, the student must become familiar with 
differentials. In the following pages, therefore, we shall fre- 
quently give the differential formulae along with the derivative 
formulae. 



CHAPTER IX 
THE DERIVATIVE AS VELOCITY 

76. Definition of Velocity. Let a body be in motion along a 
path that may be either straight or curved. If equal distances be 
traversed in equal times, the motion is denned to be uniform, 
and the distance traversed in a unit of time (second, minute, 
hour, . . . ) is defined to be the speed or velocity of the motion per 
unit of time. Plainly, 

distance 



velocity = 



time 






and this ratio is constant. 

If on the other hand the distances traversed in equal times are 

unequal, the motion is said to be non-uniform. For example, a 

body falling to the earth falls different distances in different 

seconds, or in different tenths of a second, or in different hundredths 

of a second, etc., and its motion is non-uniform. In the case of 

dists npe 

non-uniform motion, we define the ratio, ——. , to be the mean 

time 

or average speed or velocity per unit of time* Mean velocity is the 

mean or average distance the body moves in a unit of time, though 

this mean distance may be quite different from the actual distance 

covered in any particular unit of time during the continuance of 

the motion. For example, during the first 4 seconds of a body's 

fall to the earth, it covers a distance of 256 feet, and consequently 

has an average speed of —r- = 64 feet per second, while the 

* In the text we have used the term velocity as synonymous with speed. 
Strictly defined, however, velocity should involve the idea of direction of motion 
as well as of speed. 

103 



104 DIFFERENTIAL CALCULUS §77 

actual distances covered during the 4 successive seconds of its fall 
are respectively 

16 feet, 48 feet, 80 feet, 112 feet. 

Mean velocity is also the velocity the body would have to have 
if its motion were uniform, in order to accomplish the given dis- 
tance in the given time. 

77. Instantaneous Velocity. Mean velocity serves very well 
to describe the motion as a whole, but tells us little or nothing of 
the nature of the motion at any given instant. We seek now a 
means of describing or characterizing the motion at a particular 
instant, and shall find this in the instantaneous velocity, which we 
proceed to explain. 

Consider first the example just given, of a body falling from 
rest under the influence of the earth's attraction. If s be the 
distance in feet fallen in t seconds from the start, it is found by 
experiment that s is given by. the equation 

8= 16 £ 2 . 

We seek a property of the motion which shall be characteristic 
of it a€ a given instant, say at the end of the third second. 

When t = 3, s = 16 (3 2 ) = 144 feet. 

Now consider an interval of 0.5 second immediately following the 
third second. 

When t = 3.5, s = 16 (3.5) 2 = 196 feet. 

Then the distance fallen in this particular 0.5 second is 196 — 
144 = 52 feet, and the mean velocity during this 0.5 second is 

— = 104 feet per second. 
U.o 

Now take an interval of 0.4 second immediately following the 

third second. 

When* = 3.4, s = 16 (3.4) 2 = 184.96 feet. 

The distance fallen during this 0.4 second is 184.96 — 144 = 

40.96 



40.96 feet, and the mean velocity during the 0.4 second is 
102.4 feet per second. 



0.4 



§77 



THE DERIVATIVE AS VELOCITY 



105 



In a similar manner we find the mean velocities during intervals 
of 0.3 second, 0.2 second, 0.1 second, 0.01 second, following the 
third second, and construct the following table: 

Intervals following the third second. 

0.5 second 104 



.4 
.3 
.2 
.1 
.01 



Mean velocities. 


104. 


feet 


per second. 


102.4 


a 


It it 


100.8 


it 


it it 


99.2 


a 


it it 


97.6 


it 


n ti 


96.16 


it 


it n 



Now the mean velocity of 104 feet per second is descriptive of 
the motion during the 0.5 second following the third second, and 
so is approximately characteristic of the motion when t = 3 seconds. 
The mean velocity of 102.4 feet per second during the shorter 
interval of 0.4 second is a closer approach to being characteristic 
of the motion when t = 3 seconds. And so on down the table, 
we have a series of mean velocities during intervals which are 
shorter and shorter, and so these mean velocities approach more 
and more nearly to being characteristic of the motion when t = 
3 seconds. The limit of this series of mean values is the charac- 
teristic we are seeking, and is termed the instantaneous velocity 
when t = 3. That limit may be found as follows: 

Let At be the time interval in seconds, following t = 3 seconds. 
During the time (3 + At) seconds, the body falls 

16(3 + A£) 2 = (144 + 96 A* + 16 At 2 ) feet, 

and the distance fallen in the interval At is (96 At + 16 A^ 2 ) feet. 
Therefore the mean velocity during the interval At is 

96 A* + 16 At 2 



A* 



= (96 + 16 At) feet per second. 



Finally, the limit of this as At = is 96 feet per second. This 
limit is absolutely characteristic of the motion when t = 3 seconds, 
and is termed the instantaneous velocity at that time. Hence the 
instantaneous velocity of a falling body 3 seconds after the fall begins 
is 96 feet per second. 

Consider now the general case. Let a body start at O (either 



106 



DIFFERENTIAL CALCULUS 



§77 




t+At 



line) and reach the point A in time t, and the point B in time 
t + At. Let OA = s and AB = As, so that the body moves the 

distance As in time At. Its 
mean velocity during the time 

As 
At is -T- • When A 2 and con- 

At 

sequently As are made very 
small, -r- is descriptive of the 

motion over a very short path AB, and so is very nearly charac- 

As 
teristic of the motion at A. And the limit of -n as A£ = is 

At 

absolutely characteristic of the motion at A. This limit, which 
is D t s, we define to be the instantaneous velocity at A or at time t, 
or, more briefly, the velocity at A or at time t. We represent this 
velocity by v, so that we have 

D t s = v and -rr = v. 
at 

Expressed in words, the definition runs: 

The instantaneous velocity at the instant t is the limit of the mean 
velocity during a time interval At immediately following t, as At = 0. 
It is the derivative of the distance as to the time, or the "ti?ne deriv- 
ative^ of the distance. 

In the case of the falling body we have 

s = 16 1 2 , D t s = 32 1 = v. 
When t = 3, v = 96, as was found above. 

Example. 

A boy runs around a circular track 
of radius r with a constant speed of a 
feet per second. A light at the center 
of the track throws his shadow on a 
straight wall situated b feet from the 
center of the track. How fast does 
his shadow move along the wall? 

Solution. Let OAA' be _L to the wall, and let the boy start at A and 
reach B in time t so that arc AB = at. At the start and at time t the 




§78 THE DERIVATIVE AS VELOCITY 107 

shadow is at A' and B' respectively, so that the shadow moves the dis- 
tance s in time t. Then rd = at, 6 = - t, s = b tan 6= b tan - 1, 

r r 

n l a , a ab , at 
D t s =bsec 2 -t • - = — sec 2 — 
r r r r 

v — — sec 2 - 1, 
r r 

which is the velocity of the shadow t seconds after the boy starts from A. 

When t = 0, v = — ; when t = —-, = J> v= oo . 
r 2a 2 

78. Exercises. 

1. Find the velocity of the shadow in the example above when the 
light comes from the sun and the rays (which are parallel) strike the wall 
perpendicularly (harmonic motion). 

2. A bullet is fired directly upward with a muzzle velocity of 400 feet 
per second. Neglecting the resistance of the air, its height t seconds after 
the discharge is given by the formula 

s=400Z- 16 t 2 . 
Find its velocity 9 seconds and 16 seconds after the discharge. 
How high will it rise and what will be its velocity when it strikes the 
ground? How long will it be in the air? 

3. If the distance in feet passed over by a moving body in t seconds is 
given by the equation 

s=3t 3 +4:t +2, 

find the velocity at the end of 3 seconds, 2 seconds, | second. 

4. If the distance is given by the equation 

s = 10 cos - t, 

find the velocity when t= 0, |, 1, If, 2, 2|, 3, 3f, 4. 

Supposing the body to be moving in a straight line, draw a figure to 
explain its motion. This is a case of harmonic motion. 

5. A man 6 feet tall is walking along a level pavement at the rate of 3| 
miles an hour. A light directly behind him and 20 feet from the ground 
throws his shadow on the pavement. How fast is his shadow lengthening 
3 minutes after he has passed under the light? 

6. Suppose the light in exercise 5 to be placed 28 feet to one side of 
the pavement (and 20 feet above the'ground as before). How fast is the 
man's shadow lengthening 3 minutes after he has passed the light? 

7. A flagstaff 50 feet high stands on level ground. How fast is its 
shadow lengthening at 3 o'clock p.m? at 4 p.m.? (Neglect the influence 
of latitude and time of year.) 



108 DIFFERENTIAL CALCULUS §79 

8. Two ships start from the same place at the same time. One sails 
east at the rate of 12 miles an hour, and the other sails north at the rate of 
16 miles an hour. How fast are they separating at the end of t hours? 

9. Suppose the second ship in exercise 8 sails N.E. instead of N.; N.W. 
instead of N. 

10. Solve exercise 8 when the eastward sailing ship leaves the starting 
place (1), If hours ahead of the other; (2), 1 J hours behind the other. 

11. A point is moving around the circumference of a circle at a uniform 
rate of a feet per second. Find velocity of its projection upon a diameter. 
The motion of the projection is termed harmonic. See exercises 1 and 4. 

12. There is a circular race track 200 yards in diameter ,- 120 yards from 
the center of the track is a pole bearing an electric light 30 feet from the 
ground. A jockey is riding a horse around the track at a rate of 1 mile 
in 2 minutes 12 seconds. The jockey's head is 7f feet from the ground. 
Supposing the starting place to be the point nearest the light pole, how 
fast is the shadow of the jockey's cap receding from or approaching the 
foot of the pole 10 seconds after the start? | minute after the start? 

13. A ladder 20 feet long leans against a perpendicular wall, its foot 
being 2 feet from the wall. The foot of the ladder is drawn away from 
the wall at a uniform rate of 2 feet per second. How fast is the top of the 
ladder moving down the wall at the end of the seventh second? at the end 
of 8.9 seconds? at the end of 9 seconds? 

14. A balloon rises from the earth with a uniform velocity of 6 miles 
per hour. To an observer in the car, how fast is the visible surface of the 
earth increasing in square miles per hour at the start? How fast one hour 
after the start? Assume the earth to be a perfect sphere of radius 4000 
miles. 

79. Angular Velocity. If a body rotate about a fixed axis, 
every point, P, of the body, not in the axis, describes a circle 

whose center, C, is in the axis, and 
whose plane is perpendicular to the 
axis. CP, the radius vector of P, turns 
about C as a pivot, and if it turn 
through an angle A0 in time At, the 

ratio -r- is termed the average or mean 

angular velocity of the body during the time At. By the same 
reasoning as was used in the preceding article, we arrive at the 
idea of angular velocity of a body at a given instant, or the 




§80 THE DERIVATIVE AS VELOCITY 109 

instantaneous angular velocity at the time t, or, more briefly, the 
angular velocity at time t. This angular velocity at time t, which 
we denote by w, is clearly the time derivative of the angle described 
by the radius vector, so that 

To distinguish co from the velocity v described in the preceding 

article, the latter is termed tangential velocity. If s be the arc 

of the circle described by the point P, in the above figure, then 

s = rd, and 

v = D t s = rD t 0. 
Therefore 

v = rm. 

That is, the tangential velocity of a point of a body rotating about 
a fixed axis is the product of its angular velocity by its distance 
from the fixed axis. 

80. Velocity of Any Change of State. Thus far we have con- 
sidered velocity only in connection with the visible motion of a 
body. But there are other changes of state that admit the idea 
of velocity. For example, we speak of the rapidity or velocity 
with which a heated body is cooling, of the rapidity with which a 
color is fading, of the rapidity with which a gas is increasing in 
volume under the application of heat, and of the rapidity of a 
chemical reaction. And whenever the total amount of change 
of state can be expressed as a function of the time, the time 
derivative of the function (its derivative as to the time) is the 
rapidity or instantaneous velocity of the change of state. For 
example, sugar in the presence of certain acids undergoes a chem- 
ical change, and is converted into other substances. If x be the 
quantity of sugar so converted in time t, then D t x is the rapidity 
or velocity of the conversion at that instant. If a be the initial 
quantity of sugar, it can be shown that x is given by the formula 

(a) x = a(l - e~ kt ), 

where k is a known constant. To prove this is in fact just such a 
problem as that of Art. 72, and just as it was shown there that the 



110 DIFFERENTIAL CALCULUS §80 

amount of water left in the reservoir after a time t was ve~' ii i so 
here it can be shown by the same line of argument that the amount 
of sugar left unconverted after a time t is ae~ kt , and from this 
follows (a) directly. Differentiating (a), we have 

v = D t x = kae~ kt = k(a — x). 
Now a — x is the amount of sugar unconverted at time t, so that 
this equation tells us that the rapidity of the chemical reaction 
taking place at any time is proportional to the amount of sugar 
present at that time. 

Again, in certain chemical reactions it is known that x, the 
amount of the substance that is transformed in time t, is given by 

the equation 

x 

= akt, 

a — x 

where a is the initial amount of the substance and & is a known 
constant. Differentiating this equation, and reducing, we have 

v = D t x = k(a — x) 2 , 
which tells us that the rapidity of the reaction, that is, the number 
of units of the substance transformed per unit of time, is at a 
given instant proportional to the square of the amount of the sub- 
stance remaining untransformed at that instant.* 

The derivative of any function of any argument may be regarded 
as a ratio of velocities. Let y be a function of x. Then from the 
identity A A 

bm-r\ 
, .. Ay A *=o At 

we get hm -r* = -r— ; 

az=oA# ,. ax 

A<=0 A* 

or 

(6) D x y = *jy- and D t y = D x y D t x. 

* In neither of these examples does the chemist deal with the problem 

just as stated in the text. He finds rather, from theoretical considerations, 

the velocity of the reaction, that is, he starts with the equations Dtx = k(a — x) 

and D t x = k(a — x) 2 , and from these he derives the equations x = a(l — e~ kt ) 

x 
and = akt, by the process of integration, which will be explained later. 



Ay = 
Ax 


~ Ax 




At 


D x y=- 


_D t y 
D t x 



§§81-82 THE DERIVATIVE AS VELOCITY 111 

These equations show that D x y is the ratio of the speed with which 
y is changing to the speed with which x is changing, or that y is 
changing D x y times as fast as x. 

For example, let y be the volume and x the edge of a cube. 
Then y = x 3 , D x y = Sx 2 , and this last means that the volume of 
a cube is, at any instant, increasing in cubical units per second 
(or minute, or hour, etc.) 3 x 2 times as fast as the edge is increas- 
ing in linear units per second. When x = 2, 3 x 2 = 12. Hence, 
just at the instant when the edge of a cube is 2, its volume is in- 
creasing 12 times as fast as is the edge. 

When x and y are the coordinates of a point moving along a 
curve, the ordinate of the point is changing D x y times as fast as 
the abscissa. 

81. Expansion of a Metal Rod. If I be the length of the rod 

at temperature 0, Del is the ratio of the rate of increase in the 

length of the rod to the rate of increase in the temperature, and is 

termed the coefficient of expansion at temperature 6. It is proved 

by experiment that 

I = a + bd + cd 2 , 

where a is the length at temperature 0, and 6 and c are known 
positive constants. Then 

Del = b -f- 2 cd. 
This shows that the rod expands more rapidly at higher temper- 
atures than at lower. 

82. Acceleration. If v be the instantaneous velocity of a 
moving body, or of any other change of state, D t v is the speed with 
which this velocity is changing, and is defined to be the accelera- 
tion of the motion or of the change of state. It is usually repre- 
sented by a, so that we have 

dv d 2 s 
a = Dt v = D t > S or * = Jt = Wi - 

The acceleration is the second time derivative of the distance (amount 
of change of state) . 

In the case of a falling body we have 

8 = 16 1 2 ; v = D t s = 32 1; a = D t 2 s = 32. 



112 DIFFERENTIAL CALCULUS §83 

Hence the acceleration of gravitation is constant, and is equal to 
32 feet per second. 

In the case of angular velocity, «, D t u is the angular acceleration. 
Denoting this by /3, we have 

that is, 

The angular acceleration is the second time derivative of the angle 

described by the radius vector. 

Because 

s = rco, D t 2 s = rD t 2 u or a = r/3. 

That is, the tangential acceleration of a body rotating about a 
fixed axis is equal to the product of its angular acceleration by its 
distance from the axis of rotation. 

83. Exercises. 

1. How fast are the circumference and the area of a circle increasing 
compared with the increase of the radius? 

2. How fast are the surface and volume of a sphere increasing com- 
pared with the radius? When are surface and volume increasing at the 
same rate? 

3. If s, the distance traversed by a moving body in time t, is given by 
the equation 

s= at+bt 2 +ct*, 

find velocity and acceleration at time t. 

4. Ditto when s and t are connected by the equation 

s= a cos (6+ kt). 

5. Show that the acceleration of uniform motion is 0. 

6. What is meant by negative velocity? negative acceleration? Can 

the acceleration be J ~ I when the velocity is j _ j ? 

7. Boyle's law for gases asserts that when the temperature of a gas 
is kept constant the product of pressure, p, by volume, v, is constant; that is, 

pv = k. 

Using v as abscissa and p as ordinate, draw the curve. Calculate D v p and 
D p v and interpret them. 

8. A more accurate statement of the relation between the pressure and 



§84 THE DERIVATIVE AS VELOCITY 113 

volume of a gas at constant temperature is given by van der Waals' 

EQUATION, 

where a, b, k are constants. Determine and interpret D v p. 

84. A General Differentiation Formula. From the identity 

4fltt) = A/ ! (tt).Att 

Ax Au Ax' 

follows directly 
(A) D x f(u) = Duf(u) • D x u, 

which is a general formula for differentiation. This is equivalent 
to (6) of Art. 80 as may be seen by writing in (A) y in place of 
f(u), x in place of u, and t in place of x. All the general formulae 
for differentiation of Chapters IV and VIII come under formula (A) . 
For example, D x u n = nu n ~ 1 D x u is a special case of (A), in which 
f(u) = u n and D u f(u) = nu n ~ l . Again, D x smu = cos uD x u is a 
special case of (A), in which f(u) = smu and D u f(u) = cosu. In 
fact the student makes unconscious use of (A) nearly every time he 
performs a differentiation. To illustrate, in performing the dif- 
ferentiation 

D{x 2 + l) 3 = 3 2 + l) 2 2z, 

(A) is made use of. For, let x 2 + 1 = u, so that (x 2 -f- l) 3 = 
v? = f(u). Then by (A), D x a* = D u v? • D x u. But D u u 3 = 3 u 2 
and D x u = 2 x. Hence D x u 3 = 3u 2 -2x and D(x 2 + l) 3 = 
3 (x 2 + l) 2 2 x. It appears, then, that formula (A) is merely the 
formal statement of the steps of a differentiation. In performing 
a differentiation, however, there is seldom any need to make 
formal and conscious application of (A). 

Let us write down that form of (A) which is given in the first 
formula of (6) in Art. 80, and let us derive from this a second 
formula by writing in the first y' for y; we then have 

D t y 



(B) 



V = D * y D t x 
y » = Dx 2 y = D x y> = ^ 



114 DIFFERENTIAL CALCULUS §84 

These are varieties of (A) that will be found useful in a later 
chapter. 
Writing (A) in the differential notation, we have 

df(u) _ df(u) du 
dx du dx 

This formula is of course true, but we must warn the reader that 
it cannot be justified by cancelling the du's and writing 

df(u) = df(u)j6 
dx jkC dx 

This is incorrect, because the two du's are not in general equal, 
one being the differential of u considered as an argument, and the 
other the differential of u considered as a function. 
Similarly, we are not warranted in writing 

^ ^_ i 

and thence concluding that ~ . — = l and -j- = -j- • These for- 

dx dy dx dx 

dy 

mulse are indeed true, but on other grounds. 

Problem. Prove in two ways that D x y • D y x = 1 and D x y = -=r— • 

UyX 



CHAPTER X 



ADDITIONAL EXAMPLES IN CURVE TRACING 

85. Multiple Points. A point through which pass two 
branches of a curve is termed a double point. At a double point 
the curve has two tangents, one to each branch, and consequently 
y' has two values at such a point. If three branches, four branches, 
... of a curve pass through 
a point, that point is called a 
triple point, quadruple point, 
. . . , and at such a point y f 
has three, four, . . . values. 
The generic name for double 
point, triple point, ... is mul- 
tiple point. It is not our pur- 
pose to present here a general 
method for finding the mul- 
tiple points of a curve from its 
equation. That subject be- 
longs to a more advanced 
course in mathematics. Sometimes, however, the multiple points 
present themselves very simply, as is the case in the following 
examples and in some of the exercises. 

86. Examples. 
Example 1 . y 2 = x 2 {x + 3) . 

(a) The curve is symmetrical as to OX but not as to OY. It has no 
real points at the left of the line x = — 3. 

(b) Solving for y, we have 

y = ±x(x+ 3)2. 
and by differentiation 

3 x+2 „.„ , 3 x + 4 




Quadruple Point 



2 (3+ 3)*' 



y" = ± s 



4 0c + 3)t 



115 



116 



DIFFERENTIAL CALCULUS 



§86 



When y" = 0, x = —4, but y is imaginary and there is no flex. When 
y" = oo , x = — 3, but y" does not change sign at x = — 3, and there is no 
flex. Hence the curve has no flexes at all. 

(c) Consider now the branch for which the radical is +, 



y = x(x + 3)*, y' = 



3 x+2 



y 



3 z+4 



2(x+3)2' " 4( x+ 3)f 

When x is j ]M , 2/ is [ M j which means that this branch lies wholly 

in the first and third quadrants, y" is + for all admissible values of x 

(values greater than —3), and there- 
fore the branch is convex throughout 
its whole extent. 

When y' = 0, x= —2, y = — 2, and 
y" is +. Hence ( —2, —2) is a mini- 
mum point. When y' = oo , x = — 3, 
but y' does not change sign here and 
(—3,0) is neither a maximum nor 
minimum point. But the branch 
3, 0) and the tangent there is perpendicular to 
- 3. This branch passes through the origin with 




meets OX at the point ( 
OX, is in fact the line x = 
the slope Vs. It is the branch ABOC. 

{d) The second branch is traced in the same manner 
and derivatives are 



Its equation 



y=-x(x+S)*, y' = 



3 x+2 



2(z+3)^ 



3 3+4 

4(s+3)* 



It is readily found that this branch lies in the second and fourth quad- 
rants only, that in these quadrants y" is — and the branch concave, that 
( — 2, 2) is a maximum point, that the branch meets OX at ( —3, 0) and 
has there the same tangent, x = — 3, as does the other branch, and that, 
therefore, the two branches merge the one into the other at that point. 
This branch also passes through the origin, but with the slope — VS. This 
branch is lettered AB'OC in the figure. Since the two branches intersect 
at the origin with the slopes + V3 and — V3, the origin is a double point. 

Example 2. y 2 = (x + l) 2 (x - 4) 3 . 

(a) The curve is symmetrical about OX but not about OY. Solving 
for y, 

2/ = ±(z+l)(a;-4)f, 



from which it is seen that when x < 4, y is imaginary except when x = — 1 



§87 



EXAMPLES IN CURVE TRACING 



117 



and then y=0. That is, the curve has no point at the left of the point 
(4, 0) save the single detached point ( — 1,0), which is termed an isolated 
point. 

(b) The derivatives are 



y 



± 2 (X 



1) Vz-4, y' 



y" changes sign when x passes through 3, but 
of x at and near 3, and so there is no flex 
there. Moreover, y" does not change sign 
at x = 4, and consequently the curve has no 
flex at all. For all admissible values of x 
'+} .,/:„{ + 



J" 15 s-3 

4 yfcZ% 

is imaginary for all values 



(values ^ 4), when y is j _ j , 2/" is 

and the arc is J concave \ • In other words, 

the 1 1 J2L. J branch of the curve is every- 



where 



The two branches are 



U,0)l 



convex 
concave 
united at (4, 0) and the axis of x is tangent 
to each at that point. Because, further, 
neither branch can extend to the left of (—4, 0), this point is a cusp. 
See Art. 57, end. 

(c) Although y' can change sign at x = 1, y is imaginary for this and for 
neighboring values of x, and consequently there are no maximum or mini- 
mum points. In the figure A is the isolated point and B is the cusp. 

87. Exercises. Draw the following curves. Each of the first 
four curves has an isolated point. (The figures stand above their 
equations.) 



y 2 = x 2 (x-3). 
y 2 = x 2 (x 2 -2), 



3. y 2 

4. y* 



x 2 {2x 
x 2 (2x 



o 



5. 9?/ 2 




> 




6. a 2 y 2 = x 4 — b 2 x z . 



118 




DIFFERENTIAL CALCULUS 
Y 
X 



§87 



7. a 2 y 2 = b 2 x s — x 4 . 

9. y 4 -bhf=a 2 x 2 . 
Compare with 6. 

10. y^x^-x 2 ). 

11. y= (x 2 -S) 2 . 

12. y=x 4 (5-2x 2 ). 

13. y 2 {l + x 2 Y = 1. 
3a; 



14. ?/ 




15. The Witch of Agnesi: 



y 



a 2 +x 2 



16. y 



4 a; 2 
z 2 +l 



Show that this also is a witch. 
Y 




19. yi= x 2 {a 2 -x 2 )\ 

20. y 2 =z 2 (:r 2 -a 2 ) 3 . 
Compare with 19. This curve 

has 2 cusps and an isolated point. 

21. 4y 2 = (x 2 - l)* (4- a; 2 ). 

This curve has 4 flexes, 3 maxi- 
mum and 3 minimum points, and 
2 double points. 




8. ay=6¥-x 4 . 
17. 2/ 2 =4(3x 2 +2) 2 (l-a; 2 ) ; 




18. ?/ 2 = 0; 2 +l) 2 (2-a; 2 ) 3 . 
Y 




22. 36y 2 = (x 2 - l) 2 (7-x 2 ) 3 . 



EXAMPLES IN CURVE TRACING 



119 



23. y 2 =25x 2 (x+iy. 

24. a 2 x 2 =y 2 (a 2 -y 2 ). 

Hint. Use x', x" instead of 
y" . See exercise 17, Art. 63. 





25. ahf+b 2 x 6 = a 2 b 2 x\ 




27. The Cissoid of Diodes: 



y 



2 a — x 



Show that 



2/ = 



and y" = \ 

a a 2 




X 

26. The Catenary: 






y=~ 



This is the curve taken by a 
perfectly flexible cord suspended at 
two points. 



28. (y-2x 2 ) 2 =x\ 



29. 



x- 2 



30. y = 2 sin x + cos 2 z. 
exercise 9, Art. 63. 



See 




31. The Probability Curve: 



CHAPTER XI 
CURVES GIVEN BY PARAMETRIC EQUATIONS 

88. Parametric Equations. The two equations, 

(a) x=f(t), y = g{t), 

represent a curve. For, by giving a series of values to t, we get a 
series of pairs of values of x and y, which may be interpreted as the 
coordinates of a series of points on a curve. The variable t is 
termed a variable parameter, and equations (a) are termed para- 
metric equations of the curve. By eliminating t from the two equa- 
tions of (a) we obtain the x- and ^/-equation of the curve. 

One of the advantages of using parametric equations is that, 
in case f(t) and g(t) are rational functions of t, it is a simple 
matter to substitute values for t in fit) and g(t), and thus to obtain 
pairs of values of x and y, and so to plot the curve point by point ; 
whereas it may be difficult or even impossible to calculate pairs 
of values of x and y from the x- and ^/-equation of the curve. When 
it is possible to express x and y as rational functions of a parameter, 
the curve is termed a rational curve. 

Again, the forms of y' and y" derived from the x- and ^/-equation 
are not always suited to the determination of flexes, convex and 
concave arcs, and maximum and minimum points, whereas the 
parametric equations may enable us to determine these very 
readily. The following example illustrates these advantages and 
may serve as a model for the solution of the exercises that follow. 

89. Example. The Folium of Descartes. The equation of this 
celebrated curve is 

(a) y z - 3 axy + x z = 0. 

To calculate y for a given value of x would require the solution 

120 



§89 CURVES GIVEN BY PARAMETRIC EQUATIONS 121 

of a cubic equation, a tedious process. Moreover, on differenti- 
ating the equation as it stands, we get (see example 2, Art. 45) 
mJ _ay - x 2 and nJ , _ -2 aHy 



u if -ax * (y 2 -ax) z 

and it is plainly a matter of difficulty to determine where y' and 
y" change sign, and is consequently difficult to determine maximum 
and minimum points, flexes, etc. In particular we cannot easily 
determine the values of y' and y" at the origin, and the origin, as 
we shall see presently, is an important point on this curve. The 
parametric equations of this curve are 

3 at 3 at 2 

(b) x = TTv' y = T+$ = tx - 

This may be shown as follows: From (b) we get t = - , and on 

substituting this value for t in the first equation of (b) and reducing 
we get the x- and ^/-equation (a). 

We shall now study the curve by means of the parametric 
equations (b). 

When t =—1, x = y = oo, and the curve extends to infinity. 
Also when t = 0oroo,x = y = 0, and from this fact we conclude 
that two branches of the curve pass through the origin, which is, 
therefore, a double point or a cusp. 

Differentiating (b) as to t, we have 

3a(l-2* 3 ) _ 3at(2-t*) 

D & = /=i . a\* i D *y - 



(i + * 3 ) 2 ' w (i + * 3 ) 2 

whence, by formula (B) of Art. 84, 

_ Dty _ t(2 - g) v „_D t y' _ 2 (1-H 3 ) 4 
y D t x 1 - 2 f ' y D t x 3 a (1 - 2 Z 3 ) 3 

We now write down in order of magnitude what may be termed the 
critical values of t. These are the values of t that make x, y, D t x, 
D t y, y f , y" either zero or infinity, and also the values t = dzoo , in 
case these are not already included among those just mentioned. 
These critical values are 

1 3/- 

-oo, -1, 0, ~y=, V2, +00. 



122 



DIFFERENTIAL CALCULUS 



89 



Within each of the five intervals thus determined we study in 
detail the behavior of D t x, D t y, x, y, y' , y" , and shall thus be able 
to trace the curve in each interval. To fix ideas, a is assumed to 
be+. 

(1) As t increases from — oo to — .1, 

D t x is +, x increases from to + oo ) y" is + , arc convex. 
D t y is — , y decreases from Oto — oo ) the figure. 

(2) As t increases from — 1 to 0, 
D t x is +, x increases from — oo to 
D t y is —,y decreases from + 00 to 

1 



y" is + , arc convex, 
the figure. 



OC of 



CO of 



(3) As t increases from to 



</2 



D t x is + , x increases from to a V4 
D t y is +, y increases from to a \^2 
Y 



ORA 



C 



[afe a$I) 




r S aH 



\ 



y is + , arc convex, 
of the figure. 

Note that D t y changes from — to + 
^'afaafc) at £ = 0, which means that the origin 
is a point of minimum ordinate. Also 
~x y' = at this point. 

(4) As t increases from -^ to v^, 

D t x is — , x decreases from a v^ to a V2 ) y" is — , arc concave. 
D t y is +, y increases from a \^2 to a y/± ) AB of the figure. 

Note here that D t x changes from + to — at t = -^= which 

means that the point (a v^i, a \/2) is a point of maximum ab- 
scissa. Also y' = oo at this point. 

(5) As t increases from V2 to oo , 
D t x is — , x decreases from a \^2 to ) y" is — , arc concave. BSO 
D t y is —, y decreases from a VI to ; of the figure. 

We have now returned to the starting point. 

At t — V2, D t y changes from + to — , and therefore (a^ 
a v^i) is a point of maximum ordinate, y' = at this point. 

Moreover, as t increases from + oo to — oo , D t x changes from — 



§90 



CURVES GIVEN BY PARAMETRIC EQUATIONS 123 



to +, and therefore the origin is a point of minimum abscissa. 
y' — oo at this point. 

It is not actually necessary to note the points of maximum and 
minimum coordinates: they appear of themselves as the several 
arcs are drawn. 

Although y' changes sign at t = -ji=, this is not a point of max- 



imum or minimum y, and although y" changes sign at t 



1 
</2 



and 



at t = oo, the curve has no flexes whatever. Let the student 
explain this. 

It may be observed in passing that in this example y is a three- 
valued function of x. Let the student calculate the values that 
x and y have when t = 1 and — |(1 db V5). 

90. Geometric Interpretation of the Parameter. When the 
curve is rational and the relation y = tx holds, as in the case of 
the folium, the parameter, t, admits of a simple and interesting 
geometrical interpretation. The equation y = tx represents a 
line through the origin with the slope t. Let us find the intersec- 
tions of this line with the folium 



W 



3 axy + x 3 = 0. 



Eliminating y from these equations, we 
have 

x 2 (t*x- S'at + x) = 0, 



whence x 2 = or x = 



and 



y = or y = 



Sat 
1 + t 3 
Sat 2 
1 + t 3 ' 




The solution x 2 = 0, y = 0, means that the line y = tx meets the 
curve twice at the origin, which means that the origin is a double 
point. The other solution gives the coordinates of P, the only 
intersection outside the origin. As t varies, the line y = tx 
turns about the origin as a pivot, while the variable point P 
traverses the line and traces the curve. And in expressing the 



124 



DIFFERENTIAL CALCULUS 



§91 



x- and ^/-coordinates of this tracing point, P, in terms of the 
variable slope t, we have the parametric equations of the curve. 

91. Exercises. Draw the following curves. In every case 
derive the x- and ^/-equation from the parametric equations. Many 
of these curves have already been given by their x- and ^/-equations 
in preceding exercises. (The figures stand above their equations.) 

1. x=10(£ 2 -l), y=tx. 

2. x=t 2 +2, y=tx. 
Note the values of x and y when 

t=i V2, 





x= t 2 (t 2 - 5), y= iXt 



3. x=t 2 - 1, y= 5P(t 2 - 1). 

2t 2t 

4 - X 1-t 2 ' V 1 + t 2 ' 

5. x = tan d, y = sin d. 

6. x = tan e, y= cos 0. 





9. x=t 2 (t-S), y=tx. 



7. x= 10(t*-t), 



tx. 



10. x 



2-P 



, y= tx. 



91 



CURVES GIVEN BY PARAMETRIC EQUATIONS 

20. The Catenary: 

x=a\ogt, y=l(t + f) 



125 





11. x=t 2 , y=t\t+2). See Art. 
87, exercise 28. 

12. x = a sin 0, y= b sin 2 cos 0. 

13. x = a sin cos 6, y = a sin 0. 



14. x 



bt 2 



= tx. 



1+^ 5 

This curve has two cusps at the 
origin. 

15. The Ellipse: 

x = a cos 0, y=b sin 0. 

16. The Hyperbola: 

x = a sec 0, ?/ = 6 tan 0. 

17. TheAstroid: 

x= a cos 3 e, y = a sin 3 0. 

18. T 7 /^ Cissoid of Diodes : 
2 at 2 



x = 



l+P 



, y = tx, 



or x = 2 a sin 2 0, 

y=2a sin 2 tan 0. 

19. T&e Witch of Agnesi: 
x = a tan e,y= a cos 2 0. 



5' c 

21. 77ie Lemniscate of Bernoulli 



x= a cos a/cos 2 0, 
2/ = a sin Vcos 2 0. 



22. z = 



a* 4 



(t-iy 



tx. 




23. x= l-t 2 ,y= (t-l)(l-t*)K 



OA 2 a(l - J 2 ) 

or x = a (1 + cos 0) cos 0, 
?/ = a sin cos 0. 



126 



DIFFERENTIAL CALCULUS 



§92 




27. The E volute of the Lemniscate . 
2 a cos 3 



25 - *=(rnr ^= te > 

or x = a (1 + cos 0) sin 0, 
?/ = a sin 2 0. 



3 Vcos 2 

2 a sin 3 
2/ = — — • 

3 a/cos 2 

28. z=^(l-cos0), 

54 
7/ = — - (1 — cos 0) sin 0. 
4a 



26. 3 = 



4ai 2 



te, 



1-M 4 
or x = 2 a sin 2 0, 

?/ = 2 a sin a/2 sin 2 0. 




29. x = a sin 6, y= b cos 3 0. 



92. Derivation of Parametric Equations. In Art. 90 we ob- 
tained rational parametric equations of the folium of Descartes 
by substituting y = tx in the x- and ^/-equation of the curve. It 
is not always possible to derive rational parametric equations in 
this way even when the curve is rational. For example, on mak- 
ing this substitution in 



a 2 (t 2 - 1) 



(1) 


a 2 (y 2 - x 2 ) = y\ 


we get 


a 2 (t 2 x 2 - x 2 ) = t 4 x* and x 2 


whence 




(a) 


a Vt 2 - 1 

x = =*= , 2 » y = 



aVi 2 



1 



These are parametric equations, to be sure, but they are not 
rational, and are not suited to a study of the curve. And yet the 
curve is rational, as can be shown by substituting in equation (1) 
1 + t 2 



y = 



(b) 



1-t 2 



x. There results, after easy reductions, 



x = 



2 at(l - t 2 ) 
(1 + t 2 ) 2 



2 at 
1 + t 2 



§92 CURVES GIVEN BY PARAMETRIC EQUATIONS 127 

Simpler but irrational parametric expressions for x and y may be 
derived from (1) by the substitution y = x sec 6. This gives, after 
reduction, the parametric equations 

(c) x = a sin 6 cos 6, y = a sin 0* 

Sometimes still other substitutions will prove effective, but no 
general rule can be given here for deriving parametric equations 
from the x- and ^/-equation of the curve. Further discussion of 
the subject of the parametric representation of curves, includ- 
ing the subject of rational curves, belongs to a more advanced 
course in mathematics. 

Problem 1. Show that any rational function of one or more of the 
trigonometric functions 

sin 0, cos 0, tan 0, cot 6, sec 0, esc 

can be expressed as a rational function of a single parameter t where 
t= tan|0. 

Problem 2. Point out the curves of Art. 91 that are rational. 

Problem 3. Observe that in each of the exercises of Art. 91 in which 
y = tx, the origin is a multiple-point of order one less than the degree 
of the x- and ^/-equation of the curve. What general principle can be 
inferred from this fact ? 

* Equations (a), (b), (c) can be derived one from another: thus, the sub- 
stitution of sec for t in (a) or of tan|0 for t in (b) gives (c). 



CHAPTER XII 
CYCLOIDAL CURVES 

93. The Cycloid. This is the curve described by a fixed point 
on the circumference of a circle, as the circle rolls along a straight 
line. It is obvious that the curve consists of an unlimited number 
of equal arches. 

Let the fixed line be the axis of x, and let the origin be one of 
the points of contact of the generating point, P, with the fixed 
line. Let a be the radius of the generating circle, and let x and y 




be the coordinates of P when the circle has turned through an 
angle 0. Then 

x = ON - PM and y = a - MC. 
But ON = arc PN = ad, PM = a sin 0, MC = a cos 6. 

Therefore, 

x = a(6 — sin 6), y = a(l — cos 6), 
and these are the parametric equations of the cycloid. 

Problem 1. Determine D x y, and show that 4> = Z.PNM, and that 
therefore PN is the normal to the curve at P. From this show that the 
tangent passes through T, the highest point of the generating circle. 

a 



Show also that <t> = \ (t — 6) and that D x 2 y = - 

128 



V 1 



94 



CYCLOIDAL CURVES 



129 



Problem 2. If the origin be taken at H, the highest point of the cycloid, 
and if we write d = w+ d', show that the equations of the cycloid are 
x= a(d' +sin0')j y= a( — l+cosd'). 
Problem 3. Obtain the x- and ^/-equation of the cycloid. 

94. The Epi- and Hypo-cycloids. When a circle rolls on a 
fixed circle, a fixed point in the circumference of the rolling circle 
describes a curve termed an epicycloid or a hypocycloid, according 
as the moving circle is outside or inside the fixed circle. 

To obtain the parametric equations of these curves, we take the 
origin at the center of the fixed circle, and choose for OX a line 
through A, one of the points of coincidence of the generating 
point, P, with the fixed circle. Let a be the radius of the fixed 
circle and b that of the generating circle. 

In both figures, 0B= x, PB= y, arc AN = ad, arc PN = b<f>. 

Hence, since arc AN = arc PN, b<j> = ad and = r8. 



The Epicycloid. 




x = OD + PM = (a + 6)cos 6 + b sin PCM, 
ZPCM = - ZNCD = + 6 - |, 

sin PCM = - cos (<f> + 6) = -cos ^— 6. 



130 

Therefore 



DIFFERENTIAL CALCULUS 



Similarly 

The Hypocycloid. 



x = (a + 6) cos — 6 cos 
y = (a + 6) sin — 6 sin 



6 
a + 6 




z = OD+PM = (a -6) cos + 6sin PCAT, 
ZPCM = -</> + ZiVCZ)=-</> + 0+|, 



sin PCM = cos (0 — 0) = cos 



a — 



0. 



Therefore 



Similarly 



/ 7 \ n ■ i cl — b n 
x = (a — b) cos 0+6 cos — =- — 0. 



2/ = (a — 6) sin — 6 sin 



a — b 



§94 



(a) 



(b) 



Observe that the equations of the hypocycloid may be obtained 
from those of the epicycloid by changing 6 into — 6, and vice versa. 

If we write m = — r — in the one case and m = — =■ — in the 
b o 

other, we get the equations of the two curves in more compact 
form, viz., 



t = mcosd— cosmd, 

V 

j- = rasin0 — smra0, 



r = mcosd + cosm0, 

V 

r = msmd — sinra0. 

b 



95 



CYCLOIDAL CURVES 



131 



It is geometrically evident that each curve consists of a series of 
arches, and has a cusp at each point of contact of the generating 
point, P, with the fixed circle. When 6 is a divisor of a, the num- 
ber of arches and of cusps = r> and there are no double points. 

When b is not a divisor of a, the curve has double points. When a 
and b are commensurable, the number of arches, cusps, and double 
points is finite; and when a and b are incommensurable, they are 
all infinite in number. 

In the last case (a and b incommensurable), the ring included 
between the fixed circle and the circle concentric with this and 
of radius a + 2 b in the case of the epicycloid, and a — 2b in the 
case of the hypocycloid, is entirely covered by the curve. For it can 
be proved, and is indeed geometrically evident, that through every 
point of this ring pass two branches of the curve. 

95. Exercises. 

1. What is the hypocycloid when b = \al 
What does each curve become when 6=0? 

2. In the epicycloid show that 

D x y= tan £(^r — J0 = tan(0 + £</>), 

and that consequently the tangent at P passes through T and the normal 
through N. 

3. Prove a similar theorem concerning the hypocycloid. 

4. Write the equations of the hypocycloid when b = \a. Determine 
y' and the slopes of the cuspidal tan- 
gents. 

5. The Astroid. 
This is the hypocycloid when b = \ a. 

It is also termed the hypocycloid of jour 
cusps. Write its equations and reduce 
them to the forms 

x = a cos 3 0, y = a sin 3 0, 

and thence obtain the equation 



z 3 + y< 



al 



See Art. 91, exercise 17. 




132 



DIFFERENTIAL CALCULUS 



§96 




6. The Cardioid. This is the epicycloid 
in which b = a. Write down its equations, 
determine y' , and find maximum and 
minimum abscissas and ordinates. Find 
the slope of the tangent at the cusp, 
and at the points where the curve cuts 
OY. Calculate y" . Draw the curve. 

7. Discuss completely the hypocycloid 



8. Discuss 
when b = 2 a. 



completely the epicycloid 



96. The Involute of the Circle. When in the epicycloid b = qo , 
the rolling circle becomes a straight line, and may be regarded as 
the taut portion of a string wound 
round the fixed circle and carrying 
a pencil, P, which traces the curve 
as the string is wound off the circle. 
This special epicycloid is termed the 
involute of the circle. We obtain its 
parametric equations by determin- 
ing the limiting forms of equations 
(a), Art. 94, when b = oo . To this 
end we write equations (a) in the 
form 




x = a cos 6 + b ( cos 6 — cos 



a + b 



•) 



= a cos 



e+26sin ^ +1 ) 9sin g, 

y = a sin 6 + Msin 6 — sin — 7 — g ) 



x = a cos 6 + ad sin ( ^-=- + 1 J 



= asin0— 2 b cos 
These may be written 



sin 



ad 
26 



aS_ 
26 



aQ_ 
26 



§96 • CYCLOIDAL CURVES 133 

. ad 

cm 

y = a sin 6 — a^cosf^r + 1J0 



Sm 2 6 



ad 
26 

Now, when b = oo , 

. ad 

sin l^-r + 1 J0 = sin 6, cos (~-r + 1 )6 = cos 0, and — = 1 

26 
Hence the parametric equations of the involute of the circle are 



x = a(cos 6 + 6 sin 6), ) , n 



y = a(sin0 — 0cos 6). 



CHAPTER XIII 



CURVES GIVEN BY POLAR EQUATIONS 

97. The Tangent in Polar Coordinates. Let p, be the polar 

coordinates of the curve point P, 
and let </> be the angle between the 
radius vector and the tangent at P. 
We seek to express tan</> in terms 
of p and 0. Let the coordinates of 
Q be p + Ap and + A0. Let PB 
be drawn perpendicular to OQ. 

Then tan = 757: , P£ = p sin A0, 
and 
BQ = OQ - OB = p + Ap - p cos A0. 




Therefore, 

tan/3 



p sin A0 



sin A0 

A0 



Ap + p(l - cosA0) 



Ap 1 — cos A0 
A0 + p A0 



Now let Q approach P as a limit, and make use of I and II of 
Art. 11, and there results 

pdd 



(a) 



tan</> 



D e p 



= P D p d = 



dp 



cot0 = -^ = A, log p. 
p 



From these equations it is seen that D e p (or D P 0) serves the same 
purpose in polar coordinates that D x y does in Cartesian coordi- 
nates, viz., determines the direction of the tangent line. 

134 



» CURVES GIVEN BY POLAR EQUATIONS 135 

Problem 1. From the figure show that when the tangent is 



' 7T — 



Problem 2. From the figure show that 

~ , p+ tan d D e p pdd + tan0dp 

D x y = tan a = -^ — ^ = ^ 4 f • 

iV — p tan 6 dp— p tan d0 

Problem 3. Obtain the same expressions for D x y by differentiating 

x = p cos 0, ?/ = p sin 0. 

Problem 4. Let p be the length of the perpendicular from the pole upon 

the tangent. Show that 



V(D e p) 2 +p 2 Vd P * + p 2 dd* 
There are formulae in polar coordinates for determining convex 
and concave arcs and flexes, but these formulas are rather difficult 
to derive and to apply, and we do not give them. 

98. Curves in Polar Coordinates. When the equation of the 
curve can be brought to the form p = /(0), and when f(6) is a 
simple function, the general shape of the curve can be readily 
determined in many cases by inspection, and by calculating p for 
a few suitably chosen values of 6, and plotting these points. All 
this is explained in analytic geometry. Our formula (a) gives 
additional information. 

Example 1. The Lemniscate of Bernoulli: 
P 2 = a 2 cos 20. 
From this we get p = ± a Vcos20. 

We need consider only one branch, 

p = +aVcos20. 

p is real only when — j <0 = + ^' anc * therefore this branch lies wholly 

within the angle BOC. When 
= 0, p has its greatest value a, and 



when d]= ±2", p 



0. Hence this 



branch forms a loop having at the 
pole two tangents BB' and CC. 
Moreover, since cos 2 = cos ( —20), 




136 



DIFFERENTIAL CALCULUS 



this branch is symmetrical about the initial line OA. By plotting a few 
points in the angle BOA, the curve can be drawn with considerable 
accuracy. The other branch p= — Vcos 2 is of the same size and shape 

and lies in the angle B'OC. The curve 
has a double point at the pole and each 
branch has a flex there. 

The curve has now been drawn with- 
out any aid from Calculus. Some of 
the foregoing conclusions, however, are 
not quite warranted. For example, we 
are not yet quite sure that the lines BB' 
and CC are actually tangents, and we are 
not at all sure that the curve is rounded 
at the points A and A', as shown in the figure on page 135. It may have 
cusps at these points, with the line AA' for cuspidal tangents, as in the 
accompanying figure. To settle these doubts, we make use of formula 




(a). Differentiating p- 



aVcos 2 0, we have 
n a sin 2d 



V cos 2 



whence 



tan <t> = ■—- = — cot 2 0, and therefore 4> = =b 



2 0. 



When = ± -, <t> = ir or 0, and this means that the tangent at the pole 

coincides with and has the same direction as the radius vector there, that 

is, that the lines BB' and CC are tangents at the pole. Since the curve 

crosses its tangent at 0, this point is a flex on each branch. When 

q 
= 0, <f> = - , and when = ir, <f> = -~ , which means that the tangents at 

A and A' are perpendicular to AA' . Hence the curve is rounded at A 
and A' as shown in the first figure. 

From the relation <t> = ^ + 20 it follows that, to draw the tangent to 

the curve at a given point of it, it is only necessary to let fall from that 
point a perpendicular upon a line through which makes an angle 3 
with AA'. 

It has not been shown that the 
curve may not have undulations 
as in the accompanying figure. A 
rigorous proof that such is not the 

case we do not give, but the fact may be rendered extremely probable by 
plotting a large number of points. 




99 



CURVES GIVEN BY POLAR EQUATIONS 



137 



Problem. A double tangent to a curve is a line that is tangent at two 
points. Find the points of the lemniscate where the tangent is parallel 
to AA' } and show that the tangents at these points are double tangents. 

Example 2. p = asin^0. 

The greatest value p can have is a, and therefore the curve lies entirely 
within a circle whose center is the pole 
and radius a. As varies from to 
5 7T, 1 varies from to ic, and P first in- 
creases from to a,(d = ifh an d then 

decreases to again (0 = 5ir). When 
is — and varies from to — 5tt, the 
same curve is traced but in the opposite 
direction. The student should trace the 
curve by plotting the points for which 




= 0, 



3jr 

4 ' 



using a table of natural sines to calculate p. A better way is to use polar- 
coordinate paper and the methods explained in analytic geometry. 

Problem. Determine <t> at the points where the curve of example 2 cuts 
OX and OY. Ans. Two of the values of <f> are 85° 17' and 81° 44'. 



99. Exercises. Draw the following curves, calculating 4> for 
important points: 

1. p = a sin 20. 
The figure shows but one loop; 

draw the others. Find the coor- 
dinates of the points where the 
tangents are parallel to the axis 
of the loop, and thus find the 
greatest width of the loop. 

Ans. to last, .544 a. 

2. p = a sin 2 d. 
Find width of each loop at 

widest part. Ans. .77 a. 

3. p = asin^0. 
This curve has two double 

Ans. .544 a. 




points. Find width of each small loop. 



138 



DIFFERENTIAL CALCULUS 



99 




4. p = a esc $9. 
This curve has two branches and two double points. 

5. p = a sin 3 0. 

This curve has three equal loops and the 
pole is a triple point. 

6. p = a sin \d. 
Find the angles at which the curve cuts the 

axes and the width of each loop. 

7. p = acsc^0. 
Observe that the radii vectores of exercises 

6 and 7 are reciprocals one of the other.' 

8. p = asin|0. 

9. The Cardioid: p = 2 a(l — cos0). 

Show that the length of a chord through the pole is constant. Show 
that <f> = 1 0, and that therefore 
tangents at the extremities of a 
chord through the pole are per- 
pendicular to each other. Find 
the points of maximum and 
minimum ordinates. Determine 
a simple geometrical construc- 
tion for this curve. 

10. The Limacon of Pascal: 

p = b — a cos 6. 

There are four cases to be 
carefully considered, 

b <a,b = a, a<b <2a,b = 2a. 

For each case draw two concentric circles of radii a and b and with the 
center at the pole, and thus get a simple geometric construction of the 
curve. 

Show that the chord through the pole is of constant length. Find the 
points of contact of the double tangents when they exist. 

ii 2m 

11. p = 

1 — cos e 

12. p = a sinf 0. 

The curve has eight double points and the pole is a multiple point. 

13. p — a sin 1 0. 

The curve has six double points and the pole is a multiple point. 




99 



CURVES GIVEN BY POLAR EQUATIONS 



139 




14. p = atanf 0. 

15. p = asin 3 0. . 

16. p = 2 a cos cos 20. 

This curve has three loops and 
a triple point. See exercise 24, 
Art. 91. 



17. The Spiral of Archimedes: p = ad. 
Let a line revolve about a fixed 

point with constant angular velocity, 
while a point, P, traverses the line 
with constant velocity along the line. 
P traces the curve. Prove this. In 

plotting the curve let a = -, that 

a 

is, plot the curve p = -. The curve 

has two branches, one for + values 
of 6 and one for — values. How 
many double points has it? 

18. The Hyperbolic Spiral: P = -• 

6 

In plotting, use the equation p = - . 



19. The Logarithmic or Equiangular Spiral: p = ae u . 

Show that tan <t> is constant, and 
that therefore the curve cuts all its 
radii vectores at the same angle. 
Hence the name "equiangular." 





The figure is drawn for &=H% 



CHAPTER XIV 



THE DERIVATIVE OF THE ARC. THE METHOD OF 
INFINITESIMALS 

100. A Theorem of Geometry. Let PQ be an arc of any 

curve, and suppose this arc to be everywhere concave towards its 
chord. We shall prove 

arcPQ 



lira 



1. 



q=p chord PQ 
Draw PT tangent at P, and QT perpendicular to PT. 



We sup- 



pose Q taken near enough to P so 
that QT does not meet the arc in any 
point other than Q. LetZ QPT = a. 
Then 
PT + TQ> arc PQ > chord PQ. 

The first inequality is a case of the 
principle of plane geometry that the 
length of any arc that is everywhere 
concave towards its chord is less 
than the length of any broken line 
inclosing it and having the same extremities. The second in- 
equality holds because a straight line is the shortest distance 
between two points. 

Now 

PT = PQ cos a and TQ = PQ sin <*, 
whence 




cos a + sin a > 



arc PQ 



> 1. 



chord PQ 

As Q =P, a = 0, sin a = 0, cos a — 1, and consequently 

arc PQ 



(a) 



q±p chord PQ 
140 



= 1. 



Q. E. D. 



§101 



THE DERIVATIVE OF THE ARC 



141 



sin $ 



1, which was proved in Art. 11, is obvi- 



The theorem lim 

0=0 9 

ously a special case of this theorem (a). 

101. The Derivative of the Arc: Cartesian Coordinates. 

Let P and Q be any two points of a curve, and let their coordinates 
be x, y and x + Ax, y + Ay. 
Let s be the length of the arc 
AP measured from any con- 
venient point A. Then s is a 
function of x, and arc PQ is 
the increment of s due to the 
increment Ax of x. We there- 
fore set arc PQ = As. Then 

lim -T— = D x s, and we seek to 

Ax=oA£ 

express D x s in terms of D x y. 

Let c be the length of the 

chord PQ. We may suppose 

Q taken at the start so near 

P that arc PQ is everywhere 

concave towards its chord. Then theorem (a) of the preceding 

article applies, and we have 

As 




lim — 
q=p c 



1, and lim 

Ax=0 C 



1. 



Therefore by the principle of Art. 10 
As 



— = 1 + e and As = (1 + e)c, 



where e is infinitesimal. From the figure c = V(A#) 2 + (Ay) 2 . 
Hence 

(1) As = (1 + e)V(Axy+(b y y and ^ - (1+ *)\Jl + (||J- 

Passing to limits, we have the important formula 

I 



2>^= Vl-f-(J^) 2 = Vl + 



V 



142 DIFFERENTIAL CALCULUS §101 

In the differential notation I has the forms 



(b) ds = Vdx* + dy 2 . 

Regarding x, y, and s as functions of a parameter t, we may divide 
the first equation of (1) by At, and then, on taking limits, we have 

II D t s = V(I> t x.) 2 -\-(I> t y)2, 

sl formula which reduces to I when t = x. 

Writing II in the differential notation, we have 

, ds 

dt 



-A%' + (W 



In this formula t may be x, y, s, or any fourth variable. Clearing 
IF of fractions, we have 
(2) ds = Vdvc 2 + dy 2 , 

which is I'(b) again, and is here proved to hold whatever the 
independent variable may be, whether x, y, s, or any fourth vari- 
able. 

It was shown in Art. 73 that the differential of the ordinate of a 
curve is the increment of the ordinate of the tangent, that is, that 
dy = BT. Hence from the figure, p. 141, and from (2), it follows 

that 

ds = PT, 

or, the differential of the arc is the increment of the tangent line. 
Therefore, the differentials dx, dy, ds are represented geometri- 
cally by the sides PB, BT, PT of the right triangle PBT. 
Since D x y = tan 6, it follows from I that 



D x s = v 1 + tan 2 6 = sec 0, 
whence 

1 . D x y 

cos = ^— , sin = ^— • 
D x s ' D x s 

By the problem and formula B of Art. 84, yr— = D 8 x, and -^ = D 8 y. 

■L/ X S L) X S 

Consequently 

III D s oc = cosQ, D s y = smO, 



§102 



THE DERIVATIVE OF THE ARC 



143 



and, in differential notation, 



III' (a) 



dx 

~r = cos 0, 

ds 



± = smd, 



(b) dx = cos 8 ds, dy = sin 9 ds. 

These equations may be derived directly from the figure if we 
remember that the sides of triangle PBT are dx, dy, ds. 

Problem. Derive III' directly from the triangle PBQ. 

In Art. 84 it was shown that D x y = -=p = tan 0. 

Then by II of the present article 
D t s 



D x s = 



whence 

IV (a) 



D,x 



Vl + tan 2 = sec 0, 



D t x = cos 6 D t s, D t y = tan 6 D t x = sin D t s. 



rw\ dx 

(b) It 



n ds 

COS 6 -r, , 

dt 



dy . Q ds 
dt dt 



Clearing IV (b) of fractions, we have a proof that III' (b) hold 
true whatever the independent variable may be. 

102. The Derivative of 
the Arc: Polar Coordinates. 

Let the coordinates of P and ^ 

Q be 6, p and 6 + A0, p+Ap. 

We seek to express D d s in 

terms of p and 6. As in the 

preceding article, we have 

As = (1 +e)c. We now draw 

PB perpendicular to OQ, and 

with as center and p as 

radius we strike the circular 

arc PR. Then 

c 2 = PB 2 + BQ*. 
Now PB = psinA0 and BQ = OQ - OB = p + Ap - pcos A0 

= Ap + p(l -cosA0). 

Hence 

c 2 = p 2 sin 2 A0 + [Ap + p(l - cos A0)] 2 , 




144 DIFFERENTIAL CALCULUS §103 

and 

As n±u/ ,/ sinAfly [A P 1 -cosAfl ] 2 

Taking limits and making use of I and II of Art. 11, we have 



V (a) n 6 s = Vf + (2> eP ) 2 . 

From this formula and from Art. 97 (a) and from the figure follow 

without much difficulty 

(b) cos$ =^=D s p, sinty =^~ s = ?D S Q; 

The corresponding differential formula? are 

(b) dp = cos ()> ds, p dQ = sin (j> ds; p = P 2 ~ r - ,pds= p 2 dQ. 

CIS 

Problem. Prove V(a) by differentiating as to d the equations 
x = p cos 6, y = p sin 6 
and substituting the results in II of Art. 101. 

103. The Method of Infinitesimals. Problems like those of 
the preceding article may be solved by the aid of the following 
theorem of infinitesimals: 

Theorem. In taking the limit of the ratio of two infinitesimals, 
each or either may be replaced by an infinitesimal whose ratio to the 
one replaced has the limit unity. 

In symbolic language this theorem runs 

Lim- = lim— n provided Urn — = 1 and Urn—, = 1, 
P (3 a p 

a, P, a', P' being infinitesimal. 

a 8 

Proof. Since lim -7 = 1 and lim —. = 1, by the principle of 
a P 

a (3 

Art. 10, — ■, = 1 + e and ~ = 1 + yj, where e and yj are infinitesimals. 
a p 

From these equations we get 

« = «'(!+*), /3 = /3'(l+Y]). 



§103 
Therefore 



THE METHOD OF INFINITESIMALS 

a a 1 + € 



145 



Let us apply this theorem 
to the determination of I and 
III, Art. 101. From the figure 
we have 



and lim - = lim -, 



(1) 



= VAx 2 +Ay 2 , 



Ax-\ 1+ [ax)' 



Ax 
c 

Ay 



cos a, 



= sin a. 



Since lim — = 1. it follows 
c 

from our theorem that in tak- 
ing the limits in (1) we may replace c by As and write 




(2) 



Ay 



Ax 

lim -r— = lim cos a, 
As 



lim 



As 



lim sin a. 



Therefore, since lim a = 6, 

(3) D x s = Vl +{D x y) 2 , D 8 x = cos 0, D 8 y = sin d, 

which are formulae I and III of Art. 101. 

Plainly we shall arrive at results 
(3) if in the foregoing argument 
we everywhere regard As as exactly 
equivalent to c, that is, if we as- 
sume the arc As to be a straight 
line coincident with its chord c. 
The triangle PBQ of the adjacent 
figure is then a right triangle whose 
(rectilinear) sides are Ax, Ay, As. 
From this triangle we get 



Y 












Q 




I 




Ay 
B 




Ax- 









X 











146 



DIFFERENTIAL CALCULUS 



103 



(4) 



As = VAx 2 + Ay 2 
Ax 



Ax V 



Ax 
Ay 



-a' 



= cos a, -jf- = sin a. 

' As 



Taking limits, we have equations (3) as before. 

Equations (4) are not accurately true: the true equations are 
(1) and (2) above. Moreover, the use of the inaccurate equations 
(4) is permissible, not because, when the triangle PBQ is very small, 
As cannot be distinguished by the eye from a right line, and equations 
(4) seem then very close approximations to the truth, but because, by 
virtue of our theorem of infinitesimals, equations (4) lead to results 
that are absolutely true. 

We may now go a step farther in the substitution of one infin- 



itesimal for another. For it can be shown that lim 



df{x) 



= 1 



(let the student give the proof), and therefore the increment of a 

function can be replaced by its dif- 
ferential in that kind of problem 
to which our theorem of infinitesi- 
mals applies. Hence in the first 
figure of this article Ay may be 
replaced by dy and As by ds. That 
is, the triangle PBT may be re- 
garded as coincident with the tri- 
angle PBQ. Then from the accompanying figure we may write 
down at once 




ds = Vdx 2 + 



dx 

ds 



cos 6, 



dy 



• o dy 

sin d, -j- 
dx 



tan0. 



From the foregoing illustrations it is obvious that there is advan- 
tage in putting our theorem of infinitesimals into the following 
geometrical form: 

In any geometrical problem which involves the taking of the limit 
of the ratio of the lengths of infinitesimal lines, either straight or 
curved, any such line may be replaced by any other infinitesimal line 
(straight or curved), provided the limit of their ratio is unity. 



103 



THE METHOD OF INFINITESIMALS 



147 



Let us apply this principle to obtain the formula? of Art. 102. 
From the figure of that article it is plain that 



PB = p sin A0, PR = pA0, lim 



PB 



, . p sin A 6 
hm ^r- = 1. 



PR p&d 

Therefore PB may be replaced by PR. Again, 

r>^ a , /, a m ^^ a BQ „ 1— cosA0 A# 
BQ= Ap + p(l-cosA0), RQ = A P , ^| = 1 + p ^ — 



whence lim 



RQ 



1, and BQ may be replaced by RQ. 



We already know that c may be 
replaced by As. Hence we may re- 
gard the lines PB, BQ, and c as 
replaced by the lines PR( = pA#), 
#Q(= Ap), and As. Therefore the 
(curvilinear) triangle PRQ (see neigh- 
boring figure) may be conceived as 
replacing or coinciding with the rec- 
tilinear triangle PBQ, and conse- 
quently to be right angled at R. We 
have then 




(5) As = V( p A0) 2 + (Ap) 2 , 

pA0 



= sin a, -r 1 - 
As ' As 



A<? V' 
Ap 



<W 



P 2 + 



COSa. 



Taking limits and noting that lim a = </>, we have 

(6) D e s = V P 2 +(Dep) 2 , pD s 6 = sin <f>, D 8 p = cos <f>, 

[q which are formula? V (a) and (b) of 

Art. 102. 
ds 

Equations (5) are of course not 

exact, but they lead to results (6) 
which are absolutely true. 

Here, too, we may write differen- 
tials in place of increments, and 
then, by mere inspection of the 




148 DIFFERENTIAL CALCULUS §104 

figure, can write down at once the formulae 

ds = Vdp 2 + (pdd) 2 , -j- = coscf), -p = sin</>, ^- = tan0. 

These formulae do not need to be retained in memory, since they 
are so readily derived from the figure. 

Examples. 

1. To determine D x s for the parabola y 1 = 2mx. 

Differentiating the equation, yy' = m, y' = — ; then 



1 + V ~ y 2 ~ 2x " - DxS ~ 7—~\~ 



2x 

2 Til 

2. To determine D e s for the parabola p = 



Differentiating, p = 



1 — cos 
2 m sin0 , , ,„ 8 m 2 



(1 — cos 0) 2 (1 — cos 0) 3 m 

104. Exercises. 

1. Find D x s for the circle. 

2. Find Z^s for the astroid, x* + 2/3 = a$. 

3. Find Z) s for the astroid, x = a cos 3 6, y — a sin 3 0. 

4. Find D^s and D e s for the cycloid, x = a(0 — sin 0), ?/ = a{\ — cos 0). 

5. Find D e s for the lemniscate, p 2 = a 2 cos 2 0. 

6. Find D s for the cardioid, p = 2 a(l — cos 0). 

n / - -\ 

7. Find D z s for the catenary, y = ^[e a -\- e ° J. 

8. Find D#s /or ^e epicycloid, and for the hypocycloid. 
Use the equation at the foot of page 130. 




CHAPTER XV 
SIMPLE FORMULA OF KINEMATICS 

105. Resolution of Velocities and Accelerations. Let a mov- 
ing body have at P an (instantaneous) velocity vi along the line 
PM , while the line PM has at the same time an (instantaneous) 
velocity v 2 in the direction PN. 
Let PA and PB represent v\ and 
v 2 in magnitude and direction. 
If now both motions become uni- 
form at P, the body will move in p, 
unit of time along PM to A, and 
in the same unit of time the segment PA will move to the position 
BQ. Q will thus be the final position of the body. Since both 
motions are uniform, the body lies always upon the diagonal PQ, 
and has in reality reached Q by traversing the diagonal PQ in a 
unit of time. Instead, then, of regarding the body as having the 
two velocities Vi and v 2 at P, we may regard it as having a single 
velocity v, represented in magnitude and direction by PQ. v is 
made up of v\ and v 2 and is termed their resultant, and vi and v 2 
are termed the components of v. v is said to be resolved into the two 
component velocities v± and v 2 ; and it is manifest that a given 
velocity v or PQ may be resolved into two component velocities 
in an infinity of ways. We have only to construct on PQ as 
diagonal a parallelogram, and the sides of this parallelogram repre- 
sent in magnitude and direction two velocity components of v. 
Or we may simply construct any triangle of which PQ is one side; 
for the line AQ also represents in magnitude and direction the 
velocity v 2 . 

149 



150 DIFFERENTIAL CALCULUS §106 

If PM and PN are perpendicular to each other, we have from the 
accompanying figure the relations 

Vi = v cos 0, v 2 = v sin 6. 

v = Vvi 2 + v 2 2 , (a) 

v = ^icos + #2sin 0. 




Any one of these three equations 
asserts that V\ and v 2 are the rec- 
tangular components of v. 
An acceleration can be resolved like a velocity. Thus, if PQ 
in the foregoing figures represents in magnitude and direction a 
body's acceleration at P, then PA and P5 represent in magnitude 
and direction two acceleration components of PQ. If PA and PB 
are perpendicular to each other, and if we represent them by a\ 
and a 2j and their resultant acceleration PQ by a, then 
(a') (1) a = V ai 2 + a 2 2 , (2) <*i = a cos 0, a 2 = a sin 0. 

106. Resolution of Velocity along a Curve. Suppose a body 

which is in motion along a curve to be at the point P at time t 

ds 
(see first figure of Art. 103) . Its tangential velocity at P is -j, and 

dx dv 
its velocities in the direction of the axes are -r: and -~ . Now by 

dt dt J 

formuhe II' and IV of Art. 101, 

< b) (1) it = v [it) +UJ ' (2) ^ - cos ^- ^ = s,n • *• 

On comparing these equations with (a) above it appears that -r. 

dy ds 

and -j: are rectangular components of the tangential velocity -r • 

Now these three velocities are in magnitude proportional to and 
in direction identical with dx, dy, and ds, the sides of the triangle 

dx 
PBT. We may so choose PB that PB = -j , and then the three 

sides of PBT represent in magnitude and direction the tangential 

ds dx dii 

velocity -r and its components along the axes -r- and ~ • 



106 



SIMPLE FORMULA OF KINEMATICS 



151 



Again, dividing by dt formulae V of Art. 102, we have 



ds 
dt 



do 



ds 



p * =8m ** 




These equations show that p -r, and -r are rectangular velocity 

components of the tangential ve- 

ds 
locity -r. Now the direction of 

ds 

■rr is along the tangent to the path 

at P, and -=- is plainly along p. 

Consequently p -r. is along a per- 

pendicular to p at P, that is, along 
a tangent at P to the circular arc described about with p as 
radius. This follows also from Art. 79. These three velocities 
are shown in the figure in direction but not in magnitude. 

Example. A point on the circumference of a circle of radius r traverses 
the curve with a constant (tangential) velocity of k feet per second. 

Find the velocities of the projections 
of the point on two perpendicular 
lines in the plane of the circle. 

Solution. We choose for axes lines 
through the center of the circle par- 
allel to the given lines. Then the 
velocities of the point upon the given 
lines are the same as upon the axes 

of coordinates, and are -r- and -~ • 
at at 

We wish merely to express these in 

terms of k and of the coordinates of P. Since -£ = k, we have by (b), (2), 

at 




dx _ 
dt ~ 


COS0 


ds 
dt 


= k\ 


X)S0, 


dy _ 
dt " 


-- sin0 


ds _ 
dt~ 


k sin 0. 


From the figure, 


cos 


= • 


-U 


and 


sinfl 


_ ? 


Therefore 




dx _ 
dt ~ 


= - 


r 


and 


dy _ 
dt 


7 £ 

k-> 
r 







which are the results sought. 



152 DIFFERENTIAL CALCULUS §107 



Since k is constant, the angle POX is proportional to t, and we may set 
POX 
quently 



k 
4- POX = pi where n = - . Then x = r cos p.t, y = r sin ^£, and conse- 



efo 7 . . dy , . 

— = — k sin fj.t , -r- = k cos pX. 
at at 

The motion of the projections is termed harmonic. (See exercise 11, 

Art. 78.) 

107. Exercises. 

1. A point is moving along the parabola y 2 = 2mx. What must be 
its tangential velocity in order that the velocity of its projection upon 
OY shall be Vm? What will then be the velocity of its projection upon 
OX? 

2. With the focus as a pole, the polar equation of the parabola is 

m 

1 — COS0 

When the radius vector turns about the pole with a constant angular 
velocity k, what are the velocity components along the tangent and along 

the radius vector? Note what these become when d = x, ^ , 0. 

3. A bicyclist is riding at the rate of k feet per minute, and the radius 
of his wheel is a feet. Find the velocity of the projection upon the 
ground of a point on the rim of the wheel. Where is this velocity the 
greatest? the least? Does the point on the rim ever move backward? 

4. A point moves around an ellipse with a constant tangential velocity 
k. Find its velocity components parallel to the principal axes of the 
ellipse. 

In the following curves the radius vector turns about the pole with a 
constant angular velocity k. Find the velocity components along the 
radius vector and along the tangent. 

5. The logarithmic spiral, p — ae w . 

6. The spiral of Archimedes, p = ad. 

7. The lemniscate of Bernoulli, p 2 = a 2 cos 2 d. Note what the tangen- 
tial velocity becomes when e = 0, ^, ^, tt. 



CHAPTER XVI 




CURVATURE. EVOLUTES AND INVOLUTES 

108. Curvature. The curvature of an arc is, in everyday 
phrase, its deviation from a straight line. A straight line has 
everywhere the same direction, 
while a curve changes its direction 
from point to point. We may 
therefore define the total or abso- 
lute curvature of an arc to be its 
total change in direction. It is 
measured by the angle through 
which the tangent line turns as 
the point of contact moves from 
one end of the arc to the other. 

Thus the absolute curvature of the arc PQ in the figure is 
a = 6' — 6. It is evident that the absolute curvature is also 
measured by the angle between the normals at the ends of the arc. 

The mean or average curvature is the ratio of the total curvature 
to the length of the arc. It is what the total curvature of each 
linear unit of arc would be if the curvature of the arc were the 

OL 

same throughout. The mean curvature of the arc PQ is -p-~ • 

a 
Now let Q = P. The limit of r— is defined to be the actual 

curvature, or simply the curvature at the point P. 

We may write As for PQ and A0 for a, and then, if a denote the 
curvature of the curve at P, we have in accordance with our 
definition 

I e = lim ~r =D,6. 

a*=o As 

153 



154 DIFFERENTIAL CALCULUS §108 

We seek now to express a in terms of x and y. By formula 

D 6 
(B) of Art. 84, we have a = jf-' Now 6 = tan -1 D x y; therefore 

D x 6 = Y^rm-yi' And ' by *' Art ' 101 > DxS = Vl + (Z) ^ )2 * 

Therefore 

II g- J.'y _ y" f 

[1 + (D^) 2 ]I (1 + J,' 2 )* 
or in differential notation 



<r = 



(efcc 2 +di/ 2 )| ^ 



As there will be no occasion to distinguish between positive and 
negative curvature, that sign is to be given to the radical in II 
which will render a +. This formula enables us to compare the 
curvatures of a curve at different points. For example, in the 
parabola y = x 2 we have 

y f = 2x, y" = 2, 1 + y"> = 1 + 4* 2 , and .\ a = + \ x2)f 

It is at once apparent that the parabola is curved most at the 
origin [x = 0), and that, since a = as x = oo , the curve ap- 
proaches more and more nearly the form of a straight line, as we 
go out along the curve. 

Obviously the curvature of a right line is 0. Consider now the 
circle of radius r. The mean curvature of an arc PQ{ — ra) is 

a _ a _ 1 
PQ~n*~~r ) 

and therefore the actual curvature, 
which is the limit of the mean curva- 
ture, is also - ; that is, 

1 
a = — ■ 

r 

Expressing this result in words, the curvature of a circle is the same 
at every point, and is equal to the reciprocal of the radius. A circle 
may therefore be drawn having any curvature from to oo . Since 




§109 



CURVATURE. EVOLUTES AND INVOLUTES 



155 



r = -i a circle of curvature is a circle of oo radius; that is, it is 
a 

a right line, while a circle of oo curvature is a circle of radius, 

that is, a point. Hence the point and the right line are the 

extremes of curvature. When a = 1, r = 1, or the circle of unit 

curvature is the circle of unit radius. 

109. The Circle of Curvature. From what has been said it 
is plain that we may calculate the curvature of a curve at any 
given point P, and then by constructing a circle whose radius is 
the reciprocal of the curva- 
ture, shall have a circle which 
has the same curvature as the 
given curve at P. If now we 
so place this circle that it shall 
be tangent to the curve at P 
(have with the curve a common 
tangent line at P) and shall 
have its center on the concave 
side of the arc, it is termed 
the circle of curvature or the 
osculating circle of the curve 
at P. Its radius is termed the 
radius of curvature, and its 

center the center of curvature of the curve at P. We may regard 
the curve as coinciding with the circle of curvature along an 
infinitesimal arc about P, and may thus conceive the curve as 
made up of infinitesimal circular arcs. Any number of circles 
may be drawn tangent to the curve at P, but the circle of curva- 
ture is distinguished from all the others, in that it has the shape 
of the curve at P. If r be the radius of curvature at the point 
(x, y), we have, by II of the preceding article, 

o" V 

where that sign is to be given to the radical that will make r 
positive. It is now a simple matter to determine the coordinates 




156 



DIFFERENTIAL CALCULUS 



§109 



F 


c 








. 




P{n,y) 







A 


X 











of the center of curvature. Let 
C be the center of curvature of 
P. C lies on the normal at P and 
at a distance r from P. Then, 
from the figure, 

a = x — r sin 0, 

j8 = y + r cos 6. 



(a) 



Further, 

sin0 = 
Hence 
IV 



(l + y'iy (i+y f2 Y y' 



a = x 



v'{l + v' 2 ) 



P = y 



i+y 



V 

m 

Example 1. To find a, r,a, /3 for the parabola y 2 = 2 mx. 

nil- i m n „ m 2 1 , /2 y 2 + m 2 2 x + m 

Solution, y' =—,\y" = -—, 1 4- 2/ 2 = — ^ — = — 9 ^ ■ 

Then 



= y. 



Also 



(1 + y' 2 Y (y 2 + m 2 ) 2 (m + 2 x)' 
= 1 = (y 2 + m 2 ) ! _ (m + 2 a;) 3 



Substituting in IV, we get, after easy reductions, 



= 3^ 



_ VI - 



2 V2 » 



a = 3 z + m = ~ + m, = 

2 m m 2 vm 



* 2 , 



which are the coordinates of the center of curvature of the point (x, y). 

Example 2. To find a, r, a, p for the astroid 
x = a cos 3 0, y = a sin 3 0. 

Solution. D e x = — 3 a cos 2 sin 0, D y = 3 a sin 2 cos 0. 
Hence 

y'=— tan0, D e y'= — sec 2 0, ?/" = 



_1 , 1 + w'2 = -i 

3 a cos 4 sin cos 2 



COS J 



1 



and 



3 a cos 4 sin 3 a cos sin 3 «$/ 

r = - = 3 's/axy. 

<r 

1 +?/ 2 _ 3 a cos 4 sin 



ax?/ 



Further, — ^-y. = = 3 a cos 2 sin 0, 

2/" cos 2 

whence a = a cos 3 + 3 a cos 2 sin tan = a cos (1 + 2 sin 2 0). 

|8 = a sin 3 + 3 a cos 2 sin = a sin (1 + 2 cos 2 0). 



§110 



CURVATURE. EVOLUTES AND INVOLUTES 



157 



110. Evolutes and Involutes. Each point of a given curve 
has its center of curvature. The locus of these centers of curva- 
ture is termed the evolute of the given curve. The latter is termed 
the involute of its evolute. If in equations IV we express x, y, y', y" 
in terms of x alone, or of y alone, or of any other single variable, 
as may generally be done by means of the equation of the given 
curve, we shall have the parametric equations of the evolute. Thus 
in example 1 of the preceding article we have the parametric 
equations of the evolute of the parabola, the parameter being x or 
y, while in example 2 we have the parametric equations of the 
evolute of the astroid, the parameter being 6. The a and (3 
equation of the evolute may be obtained, of course, by eliminating 
the parameter from the parametric equations. 

Example 1. To find the x and y equation of the evolute of the pa- 
rabola y 2 = 2 mx, we have only to eliminate y from the parametric equa- 
tions of the evolute 



a ^ V2 1 
= — — £3, 



o = 3a; + m ) 



found in example 1 of the preceding article. 
x = - (a - m), z 3 = — (a - mf, 



— - /S, 



2\/2' 



= -V. 



Hence 2 = -— - (« - m) 3 , 



or, writing x and y in place of a and /?, we have 




y l = 



27 m 



(x — m) 3 , 



which is the equation sought. Comparing this with the curve of example 
3, Art. 57, we see that the evolute of the parabola is a semicubical pa- 
rabola. The upper branch of the evolute consists of the centers of curva- 
ture of the lower branch of the parabola, and vice versa. 

Example 2. The evolute of the astroid. 

In example 2 of the preceding article we have the parametric equations 



158 



DIFFERENTIAL CALCULUS 



§111 



of the evolute of the astroid in the form (writing x and y in place of a 
and 0) 

x = a cos (1 + 2 sin 2 0), y = a sin (1 + 2 cos 2 0). 

A better idea of the form of the curve may be had from its x and y 

/ equation, which is obtained by elim- 
inating from the foregoing equa- 
tions. This elimination may be 
performed as follows : 
x + y = a (cos + sin 0) (1 + sin 2 0), 
x — y = a (cos — sin 0) (1 — sin 2 0). 
Squaring these and noting that 
(cos ± sin 0) 2 = 1 d= sin 2 0, we have 
(x+y) 2 =a 2 (l + sin2 0) 3 , 
(x -y) 2 =a 2 (l-sin2 0) 3 , 
(x +y)i= at (1 + sin 20), 
(x — y)i = a% (1 — sin 2 0), 

whence (# + y)* + (x — y)* = 2 a 5 , 

and this is the equation of the evolute of the astroid. That this evolute 
is also an astroid can be shown by transforming its equation to a new set 

of axes making an angle j with the old axes. The formulae for trans- 




formation are 



x = 



x' -y' 



V = 



x' + 



£(* ; p) 



V2 ' ' V2 
whence x + y = V2 x' and x — y = — V2 y' . 

Hence 2^ s'f + 2* y'% = 2 a%, 

or z'i +2/'f = (2 a) f, 

and this is the equation of an astroid. In the figure the outer curve is 
the evolute. 

111. Properties of the 
Evolute. The two theorems 
that follow make clearer the 
relation between a curve and 
its evolute. 

Theorem 1. The normals 
to a curve are tangent to the evolute, each at the center of curvature 
that lies upon it. 




§111 CURVATURE. EVOLUTES AND INVOLUTES 159 

Proof. Let C be the center of curvature of P. We wish to 
prove that the normal PC (which is of course coincident with the 
radius of curvature) is tangent to the evolute at C. We shall 
prove this by showing that the tangent to the evolute at C has the 
same slope as the normal to the given curve at P. 

In Art. 109 we obtained the equations 
(a) ' a = x — r sin 0, = y + r cos 0. 

Differentiating these as to s, we have 

D a a = D 8 x — sin • D 8 r — r cos 6D a 6. 
D,/3 = D.y + cos • D a r - r sin dD a d. 
Now, by formulae III, Art. 101, and I, Art. 108, 

1 



Hence 
Whence 



D s x = cos 0, D a y = sin 6, D a 6 = a = 

D s a = cos 9 — sin 6 • D a r — cos 6 = —sin 6 D a r, 
D a p = sin + cos 6 • D a r — sin 6 = cos 6 D a r. 

D s a aH tan 

1 



But DajS is the slope of the tangent to the evolute at C, and - 

tan 

is the slope of the normal to the given curve at P. These two lines 
have therefore the same slope, and since they have the point C in 
common, they coincide throughout, which proves the theorem. 

Theorem 2. The difference in length of any two radii of curva- 
ture is equal to the length of the arc of the evolute included by them. 

Proof. Let C\ and C 2 be the centers of curvature of P\ and P 2 
respectively, and let ri and r 2 be the corresponding radii of curva- 
ture. We wish to prove that 

r 2 — n = arc CiC 2 . 
It has just been proved that 

D s a = — sin 6D s r, D s p = cos BD s r. 
Squaring and adding these equations, we have 
(D s a)> + (ZW = (D s r)\ 



160 



DIFFERENTIAL CALCULUS 



§111 




Iir? 



Now if s represent the length of an arc of the evolute, it follows 

from formula II, Art. 101, 
that 

(D 8 ay + (D 8 (3y = (D 8 s)\ 
Hence 

D 8 s = D 8 r, Ds's - D 8 r = 0, 
x D 8 (s-r) = 0. 

Now a function whose deriv- 
ative is is a constant. 
Therefore s — r = k. 

Let A be the point of the evolute from which s is measured, and 
let ACi = Si and AC 2 = s 2 . 

Then «i — r x = h, and s 2 — r 2 = fc, 

whence 
si — ri = s 2 — r 2 , and r" 2 — n = s 2 — si = arc CiC 2 . Q. E. D. 

By aid of these theorems may be devised a simple mechanism 
for tracing the involute when the evolute is given. The relation 
7*2 = ri + arc CiC 2 is one which holds for all positions of P x and 
P 2 . Thus (see figure following) 

(r) r 2 = ri + arc CiC 2 , r 3 = r x + arc CiC 3 , r 4 = n + arc CiC 4 , etc. 

On the evolute ACi . . . C 5 B as base construct a right cylinder 
(of wood or metal) and let a cord be fastened at some point B. 
Let this cord be wrapped for 
a portion of its length about 
the cylinder, the remaining 
portion being held taut, so 
that the cord shall have the 
position BC 5 . . . C\P\. Now 
let the cord be wound off the 
cylinder. The taut portion 
will be tangent to the cylinder 
at every position, and there- 
fore, by virtue of theorem 1, 
will be always normal to the involute P\P 2 . 




As the cord 



§112 CURVATURE. EVOLUTES AND INVOLUTES 161 

is wound off the cylinder, a pencil fastened to the cord at Pi will 
trace a curve, and this curve is, by virtue of theorem 2, as 
expressed by equations (r) above, the involute P1P2 . . . P5. 

Clearly the pencil might be fastened to the cord at any point 
other than P h as at Pi, Si, ... , and in unwinding the cord 
other curves would be described, as Ri . . . R&, Si . . . $5, * . . . 
The curve ACi . . . C b B is the evolute of each of these curves. 
(Why?) Hence, although a curve has but one evolute, it has an 
infinity of involutes. 

The problem of determining, from the equation of a curve, the 
equations of its involutes belongs to the Integral Calculus. 

112. Exercises. 

Find r, a, /3 of the following curves: 

1. The circle, x 2 + y 2 = a 2 . 

2. The curve, 3 y = x z at the point (1, \). 

3. The curve, y 2 — x z . 



4. The catenary, y = h e a + e »)■ 



5. The logarithmic curve, y = log x, at the point (1, 0). 

6. The exponential curve, y = e x , at the point (0, 1). 

Find r and the equations of the evolutes of the following curves: 

7. The ellipse, b 2 x 2 + a 2 y 2 = a 2 b 2 . 

8. The hyperbola, b 2 x 2 - a 2 y 2 = a 2 b 2 . 

9. The ellipse, x = a cos 0, y = b sin d. 

10. The equilateral hyperbola, 2 xy = a 2 . 

11. The astroid, X* + y* = ai. 

12. The lemniscate, 

x = a cos d a/cos 2d, y = a sin 6 Vcos2 0. 

13. The curve, 

x = a(cos + 6 sin 0), ?/ = a(sin — cos 0). 

14. In the case of the cycloid, 

x = a(0 — sin 0), y = o(l — cos 0), 

show that r = 2 V^f, find the parametric equations of the evolute, and 
show that the evolute is an equal cycloid. (See problem 2, Art. 93.) 

15. Given the cardioid, 

x = a(2 cos - cos 2 0), y = a(2 sin - sin 2 0). 



162 DIFFERENTIAL CALCULUS §112 

Find the parametric equations of its evolute, and show that it is also a 
cardioid. 

16. Use equations III of Art. 101 to prove the formula 

V(D 8 2 x) 2 + (D 2 y y = ±- 

r 

17. From the formulae 

x = p cos 6, y = p sin 
obtain the following expression for the radius of curvature in polar coordi- 
nates: 

[,« + (D eP yf 



r = 



p 2 + 2(IV) 2 -pA> 2 p 

( P W + dp*)* 
0>W + 2 dp 2 - P d 2 p) dd 



CHAPTER XVII 
THE LAW OF THE MEAN. INDETERMINATE FORMS 




113. Rolle's Theorem. If f(x) andf(x) are real, single-valued, 
and continuous throughout an interval a = x = b, and if f(a) = 
and f(b) — 0, then f'{x) = for some value of x between a and b* 

The truth of this theorem is geometrically obvious, for it means 
simply that if the graph of f(x) cuts the z-axis at a and at b, then, 
under the hypotheses that f(x) 
and f(x) are real, single- valued, 
and continuous, there must be at 
least one point M between a and 
6 at which the tangent is paral- 
lel to the x-axis. There may, of 
course, be several such points. 

Proof. The analytic proof is as follows : If f(x) is not only 
at a and b, but for all values of x between a and b, so too is f(x), 
and the theorem is proved. Excluding this trivial case, it is plain 
that as x increases from a to b, f(x), starting from the value at 
a, must, under the hypotheses, first increase and then decrease, or 
first decrease and then increase. In either case f'(x) changes sign 
somewhere between a and b, and, under the hypotheses, f(x) can 
change sign only by passing through 0. This proves the theorem. 

* fix) must also be assumed to be differentiable, that is, to possess a deriva- 
tive at every point of the interval a to 6. In other words, we must assume 
that f'{x) actually exists. It may surprise the reader to learn that in strict 
rigor it is always necessary to assume explicitly the existence of f'{x), but there 
are functions which are real, single-valued, and continuous throughout an 
interval, and which possess no derivatives whatever at any point of the inter- 
val. All the functions dealt with in this book do possess derivatives, and 
a discussion of non-differentiable functions belongs to an advanced course 
in mathematics. A full discussion of this and of kindred topics may be 
found in Pierpont's Theory of Functions of a Real Variable, 

163 



164 



DIFFERENTIAL CALCULUS 



§114 



Any value of x between a and b may be denoted by a + 6(b — a), 
where 6 is a positive proper fraction, that is, < 6 < 1. Rolle's 
theorem asserts, then, that 

f[a + d(b-a)] = 0, 
where of course the exact value of is unknown. 

There is an important extension of this theorem known as 

114. The Law of the Mean. If f(x) andf(x) are real, single- 
valued, and continuous throughout the interval a^x^b, then 

where 6 is some positive proper fraction. 

Here, too, the truth of the theorem is geometrically obvious; for, 
from the figure we have 



b — a 



= tan a = slope of secant A B, 

and the theorem merely asserts 
that under the given hypotheses 
there is at least one point M be- 
f{b)-f(a) tween A and B at which the tan- 
gent is parallel to AB. It is 
plain that when AB is the #-axis, 
we have Rolle's theorem, which 
is thus seen to be a special case of the Law of the Mean. 
Proof. The analytic proof of the law is as follows: 
f(b)-f(a) 




Let 



K, so that 



(a) f(b)-f(a)-(b-a)K = 0. 

Let (f>(x) be a new function defined by the equation 

4>(x)^f(b)-f(x)-(b-x)K. 
Then 4>'(x) = -f(x) + K. 

Now (f)(x) and 4>'{x) are real, single-valued, and continuous wher- 
ever f(x) and f'(x) are. Moreover, 0(a) = 0, by (a), and <f>(b) =0 
identically. Therefore we may apply Rolle's theorem to <j>(x) and 

write 

tf/[a -f d(b - a)] = -f'[a + 0(p -a)]+K = 0, 



§115 THE LAW OF THE MEAN 165 

whence 

(«) f ^^=f[a + e(b-a)], 

where all we know of is that < < 1. Q. E. D. 

Clearing (a) of fractions, we have 
(«') /(&) - /(a) = (b-a) f'[a + 6(b-a)]. 

By a slight change in the notation are obtained other useful forms 
of this important law. Thus, setting b — a = h, we have b = 
a + h, and 
03) f(a + h)-f(a) =hf'(a + dh). 

Setting a — x and h = Ax in (/3), and noting that f(x + h) — 
f(x) = A/(x), the Law of the Mean takes the form 

OS') . A/(x) = Axf(x + OAx). 

The geometrical meaning of (/3) and (/3') becomes obvious on 
drawing a figure. 

Let the two functions f(x) and g(x) together with their first 
derivatives satisfy the hypotheses of the Law of the Mean. Then 
by (c/) above 

/(&) -/(«) = (6 - *)/'[« + ^i(& - a)] 
and flf(6) - y(a) = (b - a) g'[a + 6 2 (b - a)]. 

By division, 

f(b)-f(a) _ f[ a + dl (b-a)] 

g(b)-g(a) g'[a + d 2 (b - a)]' 

where, in general, B\ and 6 2 are different. The query arises, Is 
it possible to find a single 6 such that, when written in the place 
of 0i and 02, the above equation still holds true? In showing that 
such a exists we are led to an extension of the Law of the Mean 
known as 

115. Cauchy's* Theorem. If f(x) and g(x) together with their 
first derivatives are real, single-valued, and continuous through- 
out the interval a ^ x ^ b, then 

, A v /(&) -/(«) f'[a + e(b-a)] 

( ' gQ>) - g(a) g'[a + 6(b - a)] 

* Pronounced Ko'shee. 



166 DIFFERENTIAL CALCULUS §116 



Proof. Let 


fib) -/(a) K 
gib) - gia) *' 


so that 




(b) Kb) ■ 


-fia)-[gib)-gia)]K = 0. 



Form arbitrarily the function <j>(x) by writing x for a in (b) so 

that 

4>(x)=fQ>)-f(x)-[g(b)-g(x)]K. 

Then 0'(z) = -/'(*) + <7'(*)#. 

0(z) and 0'(«) are real, single-valued, and continuous throughout 
the interval a to b because f(x), g(x), f'(x), and g f (x) are so. More- 
over, <f>(a) = by (b) and 4>(b) =0 identically. Hence Rolle's the- 
orem applies to <f>(x) and we have 

*T. + . (6 -«)]-<>, whence K = g±||^j, 

and the truth of the theorem follows immediately. Q. E. D. 

If g(x) = x, then g'(x) = 1, g(6) — g(a) = b — a, and (A) be- 
comes (a) , the Law of the Mean. 

116. Indeterminate Forms. If numerator and denominator 

fix) 

of *^-4 are both or both oo when x = a, the fraction takes the 

gfr) 

form - or — , and so has no value when x equals a. On the other 

hand, it may have a limit when x approaches a as a limit. We had 

i rxr ,. e 4.- sin x 1 — cos x x 2 — a 2 „ 
examples of this m the fractions , t, , of Arts. 11 

■ r x x 2 x — a 

and 13. In the early stages of the development of the Calculus, 
mathematicians did not consider that in such a case the frac- 
tion had no value whatever when x = a, but regarded the fraction 
merely as one whose value was hard to find when x = a; and when 
they had found what we now term the limit of the fraction, they 
took this to be its elusive but true value. They called the forms 

~ and — "indeterminate forms," and this designation is still in 
oo 

common use. We shall presently meet with other "indetermi- 



§116 INDETERMINATE FORMS 167 

nate forms" besides the ones just mentioned, and there is a gen- 
eral method for finding their so-called "true values," that is, for 
finding the limits of the fractions in question. This method is 
based upon the theorems of the preceding articles. 
There are several cases. 

Case 1. The Indeterminate Form - • 

f(x) 

If f(a) = and a (a) = 0, the fraction —-{ takes the form - 

g{x) 

f( x ) 
when x = a. To determine lim^-^ we make use of Cauchy's 

x±a g(x) 

theorem, Art. 115. It is assumed, of course, that f(x), g(x), and 

their first derivatives are continuous in the vicinity of x = a. 

In formula (A) we now write x for b, and, because f(a) = g(a) — 0, 

(A) takes the form 

f(x) = f[a + d(x-a)] 
g{x) g'[a + 6{x -a)] 

Taking limits, and noting that the limit of the second member of 

fix) 

this equation is the same as lim -yr4 , we have 

z = a g(x) x = a g (x) 

Provided only that/' (a) and g'(a) are not both or both oo, we 
may write I in the form 

x = a g(x) x ± a g'(x) g'(a) 

Examples. 



. rsmfli tt r sine ,. cos0 - 

1. = -. Hence, lim = lim— — = 1. 

L 6 J^=o e=o 9 0=0 1 

1 

2. rM£l ° H li m M^ = lim| = l. 
La; — 1Ji«o x=ix — 1 z =il 

3. = tt. Hence, lim = lim—- = 2 a. 

Lx — aJ x = a x = a X — a x =a 1 

f( x ) 

Suppose now that/ 7 (a) = g'{a) = 0, so that —n-\ also takes the 

g \ x ) 



168 DIFFERENTIAL CALCULUS §116 

indeterminate form - when x = a. We then apply formula I to 
determine lim -rrk • Including fix) and g"(x) in our assumption 

x=a Q \X) 

of continuity, we have by I 

r f{x) .. f"(x) 

z = a g'{x) x =a g (X) 

and therefore 

r fix) ,. /'(a) r /"(a;) 

II lim ^7—; = lim ^tH = lim J -rrri' 

x ± a g(x) x = a g (X) x =a 9 W 

The last limit is equal to ^ttV4 provided only /"(a) and g"(a) are 
not both or go . 

And, in general, if M,££> ,£W _ _ C^ » all take the 
gix) g'{x) g"{xY g^ n Ufa) 

indeterminate form » when x = a, and if /(a;), <7(z) and their first 

n derivatives are continuous in the vicinity of x = a, then, by re- 
peated application of formula I, we get 

r fix) :. fix) v fix) r f n \x) 

III \\m J -)-4r = lmr-7^ = hnr-777-7 = • • • = lim 



gix) x ± a g'ix) x ± a g"ix) *+og in) ix) 

and the last limit is equal to , . provided only/ (n) (a) and g (n) (a) 



fix) 
are not both or 00 . That is, to find the value of lim -V4 we keep 

x = a gix) 

on differentiating numerator for new numerator and denominator 

for new denominator until a fraction is reached which is not 

indeterminate when x = a. 



Example. 



x takes the form - when x = 0. 



x — sin x 



Then by I, lim — : = lim ~*~ e , and this last fraction 

z=o x — smx x =o 1 — cos a: 

has the form - when x = 0. Hence we must apply I again; 

e x + e - x -2 r e x - e~ x 

lim — - J = lim — ; , 

2=0 1 — cos x x =o smx 



§116 INDETERMINATE FORMS 169 

and this fraction also has the form - when x = 0, and another applica- 
tion of I is necessary; 

,. e x -e~ x v e x + e~ x 2 

hm — : = lim = -• 

x =o sm x x =o cos x 1 

Therefore 

lim *'-*-.- 2 * = lim e '+ C "- 2 = lim^^ = lim 6 -^-* = 2. 
x =o x — sinz 1=0 1 — cosz x =o smx x =o cos x 

Problem 1. Prove formula I by means of (a') of Art. 114, and without 

using Cauchy's theorem. 

fix) f(a) 
Problem 2. Give a simple geometric proof that lim'^-T = ^7~T, when 

x = a g(x) g'{a) 

f («) = 0(a) =0 and f(a) and a' (a) are not both or both oo . 

Case 2. The Indeterminate Form Q • 

fix) 
If f(a) = oo and g(a) = oo the fraction :L ^-( takes the form g°- 

g[x) °° 

when x = a. In this case lim —t-t is determined by formulae I, 

II, III of the preceding case. An entirely satisfactory proof that 
these formulae are applicable here is rather difficult, and we shall 
omit it.* 

logr X 

Example. — — takes the form 99. when z = 0. 

cot a; °° 

1 

Hence lim— 7— = lim — , 

x =ocota: z =o — csc 2 x 

and the last fraction takes the form §| when x = 0. 

1 

or citi* a* o 

Now — = which takes the form - when x = o. 

— csc 2 z £ 

r log x ,. sin 2 # ,. 2 sin a: cos z ~ 

hm — ^— = — hm = — hm — — = 0. 

x =o cot a; x =o x x =o 1 

As in this example it is sometimes advisable to reduce the fraction to a 

* The proofs given in many of the textbooks are fallacious. For a satis- 
factory proof see Osgood's Differential and Integral Calculus, Chapter XL 



170 DIFFERENTIAL CALCULUS §116 

less cumbersome form rather than to continue to apply formula I to the 

1 

fraction as it stands. Thus if we apply I directly to — , we shall have 

-csc 3 a; 







1 










1 


li 


m- 

=0 


X 
— CSC 2 X 




lim 

2=0 


2 


CSC 2 


X 2 

a: cot a: 



This last fraction also takes the form || when x = 0, and if we continue 
to apply I, the fractions become more and more complicated, and always 
take the form ^- when x = 0. 

The methods of. cases 1 and 2 may be summed up in the follow- 
ing 

Rule, li, when x = a, a fraction becomes indeterminate through 

taking the form - or 2g. , its limit when x = a may often be de- 
termined by simple algebraic devices. When these fail, differ- 
entiate numerator for a new numerator and denominator for a 
new denominator, and substitute a for x in the resulting fraction. 
If this last fraction is also indeterminate, apply the process again, 
and continue thus until a fraction is reached which is not inde- 
terminate. 

Case 3. The Indeterminate Form X oo . 

If /(a) = and g(a) = go , the product f(x) g(x) takes the form 
X oo when x = a. 

To determine the limit in this case, we make use of the identity 

-/ N / N f( X ) Q( X ) 

g(x) f( x ) 

When x = a these fractions take respectively the forms -and 2|, 
and their limits can be found by the rule just given. 

Example, x cot x takes the form X oo when x = 0. 

x • 1 

Hence lim x cot x = lim— : — = lim — — = 1. 

z=o z=o tan x z=osec 2 a; 



§117 INDETERMINATE FORMS 171 

117. Exercises. Determine the following limits: 



1. 


lim* 3-3 ^ -2 


X i2 x 2 — 5 x + 6 


2. 


^-5^ + 4 


S5*-3*-4 


3. 


lim xZ - Sx ~ 2 


2 =oo z 2 — 5x + 6 


4. 


r x* - 1 


^ 


r a: 2 - 1 

lim . 



19. 




20. 


ton #. 

j/=oo e v 


21. 


lim^-, 

2/=oo e y 




(a) n > 0, (b) n < 0. 


22. 


lim sec J x • log - • 
x=i 2 x 


23. 


lim 1 -^,(n>0). 



2 = 00 X 3 + 1 

6. By the methods of this 

chapter prove the theorem at the 24. lim z n log z, (n > 0) . 

end of Art. 16. ' *=o '_ '_ 

7 l im ^ 2 - 3 ^ + 4 . 25. 1im V a + 2u -V 3a 

y =i ?/ 3 + 2/ 2 + 1 «=aVa + 5w-V6a 

,. 1 — cos0 26. lime~ M logw. 

8. lim u =oo 

A '"" l-'coBg 27 ' limsinaloga. 

9. lim a-u 

* = °l-cos0 28 - lim(l~s)tan|s." 
10. lim 1 



11. lim 



e 3 
1 — cos 2 



8±0 & 



i0 ,. 1 — COS 
12. lim 



2 

X 



13. lim 

z =o sin - A z 

14. lim^ 1 ^ 

x=0 X 

n x-2 _ 

15. lim- : 



29. 


r e — sin a; 

lim • 

2=0 cos X 


30. 


,. e* — cos a: 
lim ; 

2=0 sinx 


31. 


r tana; — sin a; 


lim 

x=0 X 3 


32. 


r x z — a 3 

lim- • 

2=a x — a 


99 


*-- 

2 



16. lim 
1 



0=|tanU-|J 
STtanz * 34. lim^-^tan^. 



x=2 a; 



17. lim 



= 1 IT- OCT T Xe 

cos - 35. lim - — - 

2 2=0 sinx 2 



18. Hm-JOS-. 36. 1^ 



r 2^X 



M =ologsinw 2=0 sin 2 x 



172 DIFFERENTIAL CALCULUS §118 



37. lim *, x ■ • I - sin" 1 x 



^ocos^-x) 39j ^2 



l log X 

oo i- u — 7T .. tan _1 x 

38. lim - — — • 40. lim — : • 

u =„-tan \u — it) x ±q smx 

,118. Indeterminate Forms (Continued). 
Case 4. The Indeterminate Form oo — oo . 

If f(a) = oo, and g(a) = oo, the function f(x) — g(x) takes the 
meaningless form oo — oo when x = a, but may have a limit when 
x = a. In such a case/(#) — g(x) may sometimes be thrown into 

the form of a fraction which, when x = a, takes the form - or — 

' ' oo 

Whereupon the limit may be determined by the rule of cases 1 
and 2. 

Example, [sec x— tan x] _* = oo — oo : to determine lim [sec z— -tan x], 

X ~~ 2 ,*5 

X 2 

T . r , i ,. 1— sinx ,. — cosx n 

Lim Lsec x — tan x\ = lim ■ — = lim — : = 0. 

■k * cosx n -sinx 

X ^2 ^2 ^2 

Case 5. The Indeterminate Forms l 00 , oo°, 0°. 

The function [f(x)]°( x) may assume an indeterminate form when 
x = a. To show this, let 

u = [f(x)] g ( x \ whence logu = g(x) log /(a;). 

Now it is evident that log u and consequently u itself can become 
indeterminate, for x = a only in case the product g(a) log /(a) 
has the form X oo . The only possible ways for this to happen 
are the following: 

(a) g(a) = oo, log /(a) = 0, /(a) = 1; then u takes the form l 00 . 

(b) g(a) = 0, log/(a) = oo, /(a) =00-; then u takes the form oo°. 

(c) g(a) = 0, log /(a) =—00, /(a) = 0; then u takes the form 0°. 

To find the limit of u in any of these cases, we have 
lim log u = lim g(x) log /(a?), 

x=a x=a 

and the last limit comes under case 3, and can be put into a form 



§119 INDETERMINATE FORMS 173 

to which the rule of cases 1 and 2 applies. Suppose m to be the 
limit thus found, so that 

lim log u = m. 

Then by (tj) of Art. 20, 

log [lim u] = m. 

x=a 

Therefore 

limu = e m , that is, lim[f(x)] oix) = e m . 

x=a 2=a 

Problem. Show that the form 0°° is not indeterminate, but is always 
equal to 0. 

Examples. 

1. 1(1 + x)*] _ n = l 00 : to find lim(l + a;)*. 

z_u i=0 

Set u = (1 + &)*, whence log u = log (1 + x) • 

This fraction takes the form - when x = 0. Therefore 

lim log w = lim — — — — - = lim — — = 1. 

2=0 2 = X x =0 1 + X 

Applying (rj) of Art. 20, we have 

l 
log Km u = 1, whence lim u = lim (1 + x) x = e. 

2=0 2=t) 2 = 

2. LC 1 + x )~ x \ = = oo°: to find lim (1 + x)*. 

X °° 2 = 00 

We proceed as in example 1, and get 

lim log u = lim -**-* =lim = 0. 

Z = O0 2 = 00 X 2 = 00 1 ~T~ X 

Then, by (t]) of Art. 20, we have 

l l 

log lim (1 + x) x = 0, whence lim (1 + x) x = e° = 1. 



119. Exercises. Determine the following limits: 

l i 

1. lim (1 + x 2 ) x . 3. lim (x + cos x) x . 

2=0 2=0 

i 4. lim(sine) tanfl . 

2. limU+z 2 ) 1 . tf= ?T 



174 DIFFERENTIAL CALCULUS §119 

5. lim (cos d) tan6 . 16. lim sec 2 a cos 5 a. 



2 



»-iSin(«-|) 



Ot = 7 



a ,. tan it — 1 J_ 

6 ' lim ~~7 ~T 17. lim a*-*. 



x=l 



7. Bm0og.2y-.logy). 18> lim ^ 



i=0 

l 



I/=0 

8. lim (1 + sin e) 9 

9 'Z l0 ^ u ■ 19< I^ (c0Saj)Xa ' 
•" ™(*-2ii)« 

2 20. lim (s tan s — J sec s). 
10. lim (1 - e z )\ -i* 2 

2=0 2 

n . i im ?5g^. 91 lim «--i 

x log tan 
9 ~2 

12. lim (tans — x). 

x=0 

& 

13. lim (1 + as)*. 

x=0 

n „ ,. sinw — ucosu 

14. lim — 

u ±o (sin ur 

15. lim (a* — l)y. 

W = 00 





"i 1 ! log z 




— i T 

cos s — - 
lim 

x=0 S 


22. 


23. 


lim s sin-, (n >0). 

x — oo S 


24. 


,. log sin 

.«■ T-T.20-" 



BOOK II 
INTEGRAL CALCULUS 



CHAPTER XVIII 
FIRST PRINCIPLES OF INTEGRATION 

120. Definitions. The fundamental problem of the Differen- 
tial Calculus is to find the derivative when the function is given. 
The problem of the Integral Calculus is the inverse of this, and is 
to find the function when the derivative is given. Or, the problem 
may be stated thus : 

Given a function, f(x), to find another function of x whose 
derivative as to x is fix) , or whose differential is f(x) dx. 

The function sought is termed the integral off(x), and the process 
of finding it is termed integration. The function to be integrated 
is termed the integrand. For example, the integral of 3 x 2 is x z 
because Dx 3 = 3 x 2 or because d(x 3 ) = 3 x 2 dx. Again, the integral 
of cos x is sin x because D sin x = cos x or because d sin x = cos x dx. 

The symbol I stands for the word integral, and the integral oif(x) 

is denoted by I f(x) dx. For example, 

I 3 x 2 dx = x 3 and I cos xdx — sin x. 

There would be no particular objection to denoting the integral 

of fix) by / f(x) instead of by / f(x) dx, were it not the almost 

universal practice in elementary mathematics and its applications 
to write the differential of the argument after the integrand. We 
shall therefore write it so throughout this book. 

175 



176 INTEGRAL CALCULUS §121 

That integration is the inverse of differentiation is expressed 
by the identities 

(a) D x Jf(x)dx=f(x) and d Cf(x) dx = f(x) dx. 

(b) JD X f(x) dx = Jdf{x) ee f( x ) . 

These identities are analogous to the identities 
tan tan -1 x = x, tan -1 tan x = x, V(x + l) 2 = x + 1, log e u = u. 
By definition the integral of 3 x 2 is the function whose differential 
is 3 x 2 dx, and x 3 is plainly such a function. But so also is x 3 + C, 
where C is any constant whatever. For, d(x 3 + C) = 3 x 2 dx. 
Hence the most general integral of 3 x 2 is x 3 + C, that is, 



/« 



3x 2 dx = x 3 + C. 

So too / cos x dx = sin x + C. 

And in general, if f(x) dx be the differential of F(x), it is also the 
differential of F(x) + C, so that 



/ 



f(x) dx = F(x) + C, 

where C is an arbitrary constant which may have any value what- 
ever. It is termed the constant of integration. Because of the 
presence of this constant there is a certain indefiniteness in the 
value of an integral, and because of this indefiniteness the integral 
just defined is called the indefinite integral. There is also a definite 
integral which will be defined in Chapter XXII. Except when 
there is some particular reason for retaining it, we shall omit the 
constant of integration and shall write simply 



/• 



f(x)dx = Fix). 



121. Two General Theorems of Integration.* 

Theorem A. The integral of the algebraic sum of a finite number 
of functions is equal to the algebraic sum of their integrals. 

* The proofs of these theorems may be omitted if the teacher prefers. 



§121 FIRST PRINCIPLES OF INTEGRATION 177 

Expressed in a formula, this theorem is 
(A) j[f(x)dx±g(x)dx± . . . ]=Jf(x)dx±jg(x)dx± 

Proof. By formula IV of Art. 73, we have 

d\ I f( x ) dx zt I g(x) dx ± . . . = d j f(x) dx±d I g{x) dx ± . . . 

= f(x) dx ± g(x)dx ± ... by (a), Art. 120. 
Therefore 

Cd\ if(x)dx ±j g(x)dx ± . . . = f[f(x) dx ± g(x) dx± . . . ]. 

But by (b), Art. 120, 

I d\ J f(x)dx ± / g{x)dx ± . . . = / f(x)dx ± f g(x)dx[± . . . 

Equating the second members of the last two equations, formula 

(A) results. Q. E. D. 
Theorem B. A constant factor may be transferred from one side 

of the integral sign, I , to the other side without affecting the value of 

the integral. 

Expressed in a formula, this theorem is 

(B) fcf(x) dx = c ff(x) dx. 

N. B. This theorem is true of a constant factor only. 
Proof. By Art. 73, III, and by (a) of Art. 120, 

d \c f f(x) dx \= cd f f(x) dx = cf{x) dx. 

Therefore / d \c I f{x)dx = I cf{x)dx. 

But by (b), Art. 120, 

I die I f(x)dx = c I f{x)dx. 

On equating the second members of the last two equations we get 
formula (B). Q. E. D. 



178 INTEGRAL CALCULUS §122 

122. Fundamental Formulae of Integration. It will be proved 
later (Art. 171) that every function that satisfies certain condi- 
tions of continuity, etc., has an integral. But the integral of a 
given function is not necessarily one of the elementary functions 
that we already know, that is, is not necessarily an algebraic, 
trigonometric, circular, exponential, or logarithmic function. For 
example, Vx z + 1 has an integral, but this integral is not one 
of the functions just enumerated: it is in fact an elliptic integral. 
Further, there is no direct and infallible method of finding the 
integral of a given function, because the steps of a differentiation 
cannot in general be retraced. The method followed in integrat- 
ing a given function is this : From our acquaintance with deriva- 
tive formulae, we endeavor to recognize the given function as the 
derivative of a known function, or to bring it to a form in which 
it is so recognizable. If we fail in this we conclude either that we 
are not skillful enough to bring it to a recognizable form, or that 
the integrand is the derivative of a function with which we are 
not acquainted, as was Vx 3 + 1 above. Thus 3 x 2 and cos x 
were at once recognized as the derivatives of x 3 and sin x respec- 
tively. On the other hand, x cos x 2 dx is not immediately recog- 
nizable as the differential of a known function. But it may be 
written 

x cos x 2 dx = \ cos x 2 • 2 x dx = \ cos x 2 d(x 2 ), 

and the last form is seen at once to be the differential of J sin x 1 . 
Hence / x cos x 2 dx = \ sin x 2 . 

From the derivative formulae of Chapters IV and VIII, or from 
the differential formulae of Art. 73, we shall derive a set of inte- 
gration formulae which give the integrals of some of the simpler 
functions. Every process of integration consists in bringing the 
integrand to a form to which one of these formulae applies. For 
convenience of reference we assemble these formulae here. 



122 FIRST PRINCIPLES OF INTEGRATION 179 

Fundamental Formulae of Integration.* 
A. Algebraic Functions. 

/u n+1 
u" du = — — ■ (fails when n = —1). 
n + 1 

ir 1 du = f — = log u (the complement of I). 



II 
T „ C du 1 _.u 1 

in. / a, 2 = ^ tan " or — cor 1 -. 

du . , u 

= sin -1 - .. 
Va 2 -u 2 a <* 



IV. I , ™™ == = sin x — or —cos l — 



J u 



du 1 _, i* 

= - sec L — or esc" 



V «* 2 - a 2 a a a a 

B. Exponential Functions. 

VI. I a u du = =-^— • VII. I e u du = e tl . 

J log a J 

C. Trigonometric Functions. 

VIII. I cosudu = sinu. XI. / csc 2 udu = —cotu. 

IX. I sinudu = —cosu. XII. I secutsinudu = secu. 

X. I sec 2 udu = tanu. XIII. J cscucotudu = —cscu. 

XIV. I tsmudu = — logcosw = log sec u. 

XV. f cot udu = log sin w = — logcscw. 

* Note to the Teacher. — The student who knows the formulae of differ- 
entiation will have no difficulty in memorizing the .formulae of this article. 
And it is better that he should do so, although all these formulae are contained 
of course in the table of integrals. Very few other formulae of integration 
need to be memorized, because in the following chapter he will learn how to 
use a table of integrals, and he should use it in all the more difficult problems 
of integration. In order, however, to make intelligent use of such a table, 
the student ought first to have some practice in integrating without a table. 
The teacher may find it advisable to supplement the exercises of the present 
chapter with others of the same kind. 



180 INTEGRAL CALCULUS §123 

N. B. To make these formulae entirely general, an arbitrary 
constant of integration should be added to the second member 
of each. 

It is a sufficient verification of these formulae to show by differ- 
entiation that the derivatives of the second members are the 
integrands of the first members. In some cases, however, other 
proofs will be given. 

123. Proof and Application of Formulae I and II. 

— — = u n du \ / a»du = — — • 

n + 1/ J n + 1 

Proof of II. d log u = — \ I — =1 u" 1 du = log u. 

Examples. 

1. C(7x h + 5x%)dx = C7 x*dx+ C 5 x%dx = 7 C x h dx+ 5f x% dx, 

by theorems A and B of Art. 121. 
To the last two integrals I applies directly, and we have 

/**- 5 + 1 = 6* and 5 X3dx " T+i " i* 1 - 

Hence J (7 a; 6 + 5 x$)dx = | a: 6 + 3 xi 

by A and 5 of Art. 121. By I and II 

fx~ 5 dx - 3TZI = 1 **' J"*** = *' and / ~x = l0g * 
Hence f (— s - 2 - -^dx = 15a* - 2x - 3 log x. 

In applying formula? I . . . XV, it should be kept in mind that 
the u of the formulae stands for any function of x, and that du 
is the differential of this function. The following examples will 
make clear the significance of this remark. 

3. fVx-2dx. 



§124 FIRST PRINCIPLES OF INTEGRATION 181 

Observe that dx = d(x — 2), and that therefore we may write 

JVz-2 dx = f(x- 2)U(x - 2). 

To this last, integral, formula I applies directly if we regard u of that 
formula as standing for x — 2. Hence 

fVF=2dx = j(x - 2)U(x-2)= ( x ~^ +1 = | (a ._ 2)f. 

4. J(3z 2 -7z + 2)3(6z-7)d:e. 

We see that (6 x — 7)dx = d(S x 2 — 7 x + 2), and that therefore, if we let 
u of formula I stand for 3 x 2 — 7 z + 2, this formula will apply to this 
case, and we shall have 

J (3 x 2 - 7 z + 2)3 (6 x - 7)dx =f(Sx 2 -7x + 2)U(S x 2 -7x-\- 2) 

= f(3z 2 -7.T+2)l 

/- 6 a: - 4 z 3 , = _ r d(x* - 3 s 2 -f 2) 
Jz 4 -3z 2 + 2 * J z 4 -3x 2 +2 

= (byII)-log(z 4 -3z 2 +2). 
6. f sin a; cos zdz. 

In this case cos xdx = d sin x, and / sin x cos xdx = I sinxd sinx, 

and to this I applies if we let u stand for sin x. Hence 

C a C a eu n ( sin ^) 1+1 i • a 

I sin x cos x ax = I sin xa sin x = (by I) — = f sm 2 x. 

Or, we may proceed thus : 
/ sin x cos xdx = I cos x sin x dx = — f cos x d cos x = — J cos 2 x. 

Here we have two different results, but as they differ only by an 
additive constant (J sin 2 x = | — § cos 2 x), both results are right. 

124. Devices for Bringing the Integrand to a Form to Which 
the Fundamental Formulae Apply. These devices should be 
carefully studied, because they are of use not only in applying I 
and II, but also in applying each of the other formulae. 



182 INTEGRAL CALCULUS §124 

(a) The introduction of a constant factor. 
1. f (4 z 3 - 5)* x 2 dx = ^ f (4 x 3 - 5)* 12 x 2 dx 

'(4x 3 -5)^(4z 3 -5) 



-a/' 



n 1 (4a; 3 -5)* 1 ,. , CN 7 
= (byI) 12 J = 32(4^-5)- 

x 2 +2x ^_1 f3z 2 +6:r ^_ 1 fd (re 3 +3 x 2 +5) 

z 3 +3z 2 +5 



" J z 3 +3x 2 +5^~3j z 3 +3z 2 +5^~ 3j 



= (byII)|log(z 3 + 3:r 2 + 5) 



= logVa; 3 + 3z 2 + 5. 
(/3) Division of numerator by denominator. 

= (by I and II) z 3 -7 z+log(z 2 -l). 
Such division should always be performed when the numerator is 
of the same or higher degree than the denominator. 

(7) Expansion. 

4. f(x*-l)(x + 2)dx = f(x* + 2x*-x-2)dx 

/y*5 /v*4 /y»2 

= 5 + 2 ~ 2 ~ ** 
An integrand should not be expanded unnecessarily. For example, 

/ (x 3 — l) 2 x 2 dx maybe expanded and integrated, but it is better 

to proceed as follows: 

1 (x s - l) 3 



5. f(x 3 - l) 2 x 2 dx = I f(x 3 - l) 2 d{x 3 - 1) = 



3 3 

(x 3 - iy 
9 



The correctness of the foregoing integrations, as indeed of any 
integration, may be tested by differentiating the results. 



§125 FIRST PRINCIPLES OF INTEGRATION 183 

125. Exercises. Perform the following integrations. Test 
results by differentiation. 

1. J*(4x 3 -6x 2 + 2)dx. 17. f^npjfdp. 

2 S**- ^ Si^7T? dy - 

4 . ("WLJfc 20 . f 3 ? /-2 



/ ydy 
y % -l 2L JVl-2t/ + 7/ 3 (3?/ 2 -2)*/. 



*/ 



ydy 

^/y 2 -- 1 ' 22. J"(l - x + x 2 i 2 dx. 

1 — x , /*dx /*dx 



7 - J" X(fa = J*i-/f' 23. J|^| 



•x 2 -l 



JV" dx - 24. r4* 



Q r x°ax 
y ' J x 2 - 1 



xMx ^V^-3 



10. 



x 2 dx 



/ x*ax 
x 3 -l 



x 3 - 1 
x 2 
xdx 



2. f 

J (x 2 + 5) 3 



12 

J (a? 



25. r^ai*. 

26. ftanxsec 2 xdx. 

27. f secxtanxsecxdx. 

28. JVl — sine cos Odd. 



Vl - z /> sin cfa 



14. ^(a + 5) 3 d«. 



VI — cos 



30. J- sm 



Integrate in two ways and ex- J v 1 — 2 

plain why the results are not iden- ~ sec 2 6dd 

tical. . 3L J 4 + 3tan0* 

15. f (x 4 -x -3)^(4 x 3 -l)dx. 32. ftan0cos0d0. 

16. j (x 3 - x 2 + 1) (3 x 2 - 1) dx. 33. f(z + -)(l- ~\ dz. 



184 INTEGRAL CALCULUS §126 

126. Proof and Application of Formulae III, IV, and V. 

r> r r ttt C du 1, _,« 1 . ,U 

Proof of III. / -o— : — s =- - tan x - , or cot -1 - • 

J or + u* a a 1 a a 

d(-) 

Moreover, tan -1 - = - — cot -1 -, whence d tan -1 - = — d cot -1 - • 
a 2 a a a 

Consequently either - tan -1 - or — - cot -1 - may be taken as the 
a a a a J 

value of the integral / 2 . Observe that these two values of 

the integral differ by the constant - ■ 

z 

r> s sttt T dll . _.U _,M 

Pr ° 0/0/IV - JVa^^ =Sm = ' ° r " C0S a 
fdu == r d \a) =g . n _ l! . 

Also, since sin -1 - = - — cos -1 - , and d sm -1 - = — d cos -1 - , it 
a 2 a a a 

u u 

follows that either sin -1 - or —cos -1 - may be taken as the value 
a a 

of the integral in question. These two values differ by the con- 
du 1 _, u 1 



stant ~ * 



r> j- j> Tr C dU 1 _,W 1 _.U 

Proof of V. / — . = -sec x -, or esc 1 - • 

/» du __1 r d {al = l <Mflr i« (Art. 41, XIX.) 

J ^ V™ 2 - a 2 a ^ -\/( u \ 2 -1 a a 

1 u 

And it is readily shown that esc -1 - may also be taken as the 

value of the integral. 



§127 FIRST PRINCIPLES OF INTEGRATION 185 

Examples. 



i C x & x — 1 f 



+ 2 2 V3^ (V3^) 2 + (V2) 



(by III) —^— tan" 1 



2V6 V2 



o f rfa; _ 1 /• d(2%) 

J \/9-4z 2 " 2 J V3 2 - (2 xf 



= (by IV) - sin ! -y , or - - cos -1 y 
rf(V3 x) 



' J 2*V3z 2 -5 ~ 2 J V3z V ( V3z) 2 - (V5) 2 



= (byV)- i -sec-i^. 



127. Additional Devices. 



(5) Separation of a fractional integrand into partial fractions 
having^ the same denominator. 

r sx-2 _ • rxdx r 

= 3 f rf(4 a; 2 + 1) C 
8J 4z 2 + l J" 



z 2 + l 
d(2x) 



(2 a:) 2 + 1 



= (by II and III) | log (4 x 2 + 1) - tan" 1 2 x. 

o 

(e) When the integrand contains the trinomial ± x 2 -\- bx -{- c, 
where b and c may be + or — , the function may some- 
times be integrated after expressing the trinomial in the 
form dt(x + a) 2 + 0. 

2 - /*=£+« =/(SpTi = (by ln) u »- 1(x ~ 2 >- 
/■ (fa _ r dx _zlL C dVs(x-i) 

' J 6z-3x 2 -5 J 3(z-l) 2 +2 VdJ [V3(x-1)] 2 +(V2) 2 



186 



or 



J 6x-3x 2 -5 J 



INTEGRAL CALCULUS 

dx 



§128 



3(z-l) 2 + 2 



3 J (x- 



dx 



f 



dx 



= (by 111)-^= tan- 



dx 



0-l) 2 +f 
f V3 (s - 1) 



V2 



Vl - 2 x - 2 x 2 



V2J Vf^x 



x 2 + g) 



1 r d(x + 1) 

V2J Vj -(z + « 2 

= (byIV)4=sin- 1 ^y- 

V J V2 V3 

128. Exercises. Perform the following integrations, testing 
results by differentiation: 



1. f-^-. 

J x 2 + 4 

/ dx 
Vl -4z 2 



2 



13./ 



dj 



dx 



x 2 + S 
dg 
V2-32 2 * 

ydy 
V2F+3' 

da; 
9x 2 + 4* 
7 /» dx 

' J *V3x 2 -2 

2^2 



6. 



/ 



/zaz 
~z* + 



14 
15 
16 
17 

18 
19 
or 



3z 2 + 7 
xdx 


3z 2 ±7 
Vydy 


2/ 3 + 4 
dz 


zi+z^ 
dd 



f 
f 

f 

J e V4 e 2 - 1 

/ du 
Vl — w 



'/: 



du 



VS+^u-v? 

10. f- dx 

J x 



r ax _ 2 r dp*^) | 
' J Vx-x 2 J y/\ - ( x *y 
. c d(x ~ ?) 2 
J V\-(x-hY 
Reconcile the two final results. 
da. 



»•/: 



V9 z 4 - 4 

2ede 



V4-e 2 
l2 ' Je 2W +l' 



20. / 
22./ 



a + Va 

2x-3 

z 2 + l 
drc 



dx. 



V4 x — x 2 



§§129-131 FIRST PRINCIPLES OF INTEGRATION 187 

23. f *- 2 dx. 26. f , xdx . 

J V 4 x - x 2 J V 6 x - x 2 

94 f <%/ Write the numerator in the form 

' Jy 2 + 2y + 2' [S-(3-x)]dx. 

25 f %dx 27. f sec2 ^^ ! . 

' J bx 2 -2x + l ' J y/i— tan 2 x 



129. Proof and Application of Formulae VI and VII. 

VI. J a u du = ~ VII. I e u du = e u . 

Let the student prove these by differentiation. 
Examples. 

1." Ca**xdx = I Ca^d(3x 2 ) = -^— • 
J o ./ 6 log a 



2. f eV ^ x = 2f e ^dVx = 


= 2e^- 


130. Exercises. 




1. fa* x dx. 


7. fa bu + c du. 


2. fe* 3 x 2 dx. 


8. JV + l) 3 dz. 


J e v ' J 5*' 


9. JV+iPcMs. 


4. fe 9inx cos xdx. 

e C -ZJadz 
b ' J z 2 ' 


10 - /££■ 


ii. rit 1 ^. 

12. f rf * • 
J e» + 1 


6. J (e* + a;)cfcc. 


Divide numerator by denominator. 



131. Proof and Application of the Trigonometric Formulae 
VIII ... XV. Let the student prove VIII . . . XIII by dif- 
ferentiation. 



tan tidu = I = - / 

J cos u J 



dcosu 



cosu 
= — log cos u = log sec u. 



188 INTEGRAL CALCULUS §§132-133 

sins* 



XV. I COtudU = I : = / — r- 

J J smu J smu 

= log sin u = — log esc u. 

Let the student prove XIV from the identity 

'sec u tan u du 



I tan udu= j - 



sec u 
and XV in a similar way. 

132. Exercises. Perform the following integrations, testing 
results by differentiation: 

1. CsmZxdx. 11. f tanV^+T ^ 
J J Vx~+1 

2. Ccos(2x-l)dx. 12 f dx r dx 

J ' J cos 2 (a + bx) ' J sin 2 (a+fo) ' 

3. ftan2wdw. ^ o r sin a; da; 



b cos a; 



C tan. 2 udu. r_ 

J a 

4. fsin 3 cosddd. -,, r • 9 j 
J 14. j sin wsec 2 worn. 

r /* cos e dd 

5. I . r- • 1K /-cos 4 a: 
J sm 3 15. I . - - 

J sm 2 4a; 



dx. 
x 



J 2 2 16. f sec 3 2 w tan 2 w dw. 

J 17. j sm 2 a;da;. 

8. J*tan 2 wdw. #m£. sin 2 a; = f (1 - cos 2 a;). 

9. J*cot 2 wdu. 18. J" cos 2 a: da;. 

1Q r x 2 dx m 19 Csitfxdx. 
J cos 2 a; 3 J 

133. Miscellaneous Exercises in Integration. Test results by 
differentiation. 

1 r dx u 4 f(^ + l)»dy. 
•W2X-1 ^ 
r dx 5 f &_. 

/« /• 2da\ 

cot (a + mx) dx. 6. J 5x2 + 2a . + 1 ■ 



§133 FIRST PRINCIPLES OF INTEGRATION 189 

7. f VshTt cos t dt. 24. J sin (y 2 - 1) . y dy. 

z 3 + 1j„ „r rsec 2 ddd 



8. fl-^ dx . 25. f 

9- /5J| <**• 26. r- 

J cc 



COS 2 

10. C^—^dx. r a-4-1 
J z 3 4- 2 27. f ^LA 



z 3 + 1 „. 

27. r — — 

J 2z 2 + 3x + l 



* 3 + 2 27. | „ . , I , -dx. 

x z + x 



11. r^TT 1 !^. / F 

* +9 28. r Vl -^ ^ 

12. Jcos 2 (3z + 2)<fa. V^ 

1Q /■ cosxdx 29. flog sin w« cot wdw. 

ld ' Jl + sin 2 *' J 

re^±2y d 30. f>/f±*cfa. 
J e y + ?/ 2 ^ J f 1 - x 

15. fsin 2 z cos zdz. 31. f 5 - ~ 1 dx. 
J J 2 x 2 + 3 

16. f cos 2 z sin zdz. 32 f 5 x ~ 1 ^ 

' J VS-2x 2 



udu 



17 f uau , 

• J V^T 2 33. f-fc. 

_l f l-(l-2,) iu ^"^ 

2^ V^^" 2 * 34 f 8^+4o: + 13 rfa , 

^ 3 . ' J 4 x 2 — 4 a; + 5 

18- J v^a; + 1 x dx 

35. f 3 /~ 2 (fa. 

i, ;|^. 36 - /vgfe 



20. 



r 3xdx 37. f **** — 

oi f udu 38. f g^j 

■JVST^? *W3x-a; 2 -2 

22. f 2 <* 9 ■ 39. f , .g_ . 

23. Jtan(2/ 2 -l).2/% 40. J; 



3 x 2 - 4z + 3 



CHAPTER XIX 

INTEGRATION BY AID OF A TABLE OF INTEGRALS 
SOME SPECIAL METHODS OF INTEGRATION 

134. Explanation of the Table. Mathematicians have worked 
out the integration of a large number of functions of various 
kinds, and a collection of such known integrals, systematically 
arranged, constitutes a Table of Integrals. To use such a table in 
integrating a given function, we merely search through the table 
until we find an integrand similar to the given function, and can 
then write the integral down at once by giving proper values to 
the constants of the formula. Thus the process of integration is 
reduced to the purely mechanical process of substituting in a 
formula. The student should now provide himself with a table 
of integrals.* The following examples show in detail how to use 
the table. 

EXAMPLEl -/(4fe'- 

We search the table for a formula whose integrand is similar to that of 
our example and find that formula 31 fits the case. Setting a = 2 and 
b = 3 in this formula, we have 

r xdx _^1 r 1 . 1 ~|_ 1 + 3x 

J (2 + 3z) 3 ~ 9L 2 + 3z (2 + 3z) 2 J 9(2 + 3z) 2 ' 

Example 2. f J° x - 
J 8 — x z 

In this case we find that 51 is the formula that has an integrand similar 
to the one we seek to integrate. In our example a — 8, b = — 1, and 

& = y/|=-2. Therefore 

/xdx 1 rl , 4 + 2 x + x 2 . /77 , -i x + l"i 
s^-ek 108 (2-»)» +V3tan zvd: 

* The table constantly referred to in the text is that of Professor B. O. Peirce 
of Harvard University entitled A Short Table of Integrals (pamphlet edition). 

190 



§134 INTEGRATION BY A TABLE OF INTEGRALS 191 

Example 3 . f — 

J (1 + x - x 2 )* 

Formula 133 applies here. Setting a = 1, 6 = 1, c =— 1, and 
q = 4 ac — b 2 = — 5, we have 

/ x 2 c?x _ 6x + 2 _ /* dx 

(1 + x - x 2 ) 3 5 Vl + x - x 2 J Vl + x - x 2 

The last integral may be evaluated by the method used in example 4, 
Art. 127, or by formula 121. Making the necessary substitutions in this 
formula, we have 

/dx . _i 2 x — 1 

-——==. = sin — — — • 
Vl + x - x 2 V5 

Therefore 



/ 



x 2 dx 6x + 2 . _ t 2ar— 1 

; __ ! gJQ 

(1 + x - x 2 )* 3 5 Vl+x-x 2 a/5 



/x 2 c?x 
75-3 



(3 -2x 2 ) 3 

In this case formula 48 applies. Setting in this formula a = 3, b = — 2, 
m = 2, we have 

x 2 dx x 1 r di 



/ x 2 dx _ x 1 r 

(3 -2x 2 ) 3 ~ 8C3-2x 2 ) 2 8 J 



(3 - 2x 2 ) 3 8 (3 - 2x 2 ) 2 8 J (3 - 2x 2 ) 2 

To the last integral formula 41 applies, and after the necessary substitu- 
tions we get 

/ dx _ x , 1 r dx 

(3 - 2x 2 ) 2 ~ 6 (3 - 2x 2 ) + 6 J 3 - 2x 2 ' 

tt r x 2 dx x x 1_ r dx 

6nCe J (3-2x 2 ) 3 ~8(3-2x 2 ) 2 48(3-2x 2 ) 48J3-2x 2 ' 

Applying formula 40 to the last integral, we have 

\/3 + xV2 



r dx 1 1q V3 + x V 2 

J3-2x 2 2V6 ° g V3-xV2* 



Therefore, finally, 

r x 2 dx _ x x 1 , V3 + XA/2 

J (3 - 2 x 2 ) 3 8(3-2 x 2 ) 2 48 (3 - 2 x 2 ) 96 V6 g V3 - x V2 ' 

x(3 + 2x 2 ) 1 1 \/3 + xV2 

48(3-2x 2 ) 2 9 6 V6 ° g V3-xV2' 



192 



INTEGRAL CALCULUS 



§135 



Example 5. ( sin 3 x cos 2 x dx. 



This may be integrated by writing m = 2, n = 3 in formula 180. The 
result is 

/l 2 r 

sin 3 x cos 2 x dx = — - sin 2 # cos 3 £ + - J cos 2 z sin x dx. 

Now f cos 2 x sin x efc = — Ccos 2 xdeosx = — -cos z x. 

Therefore 

/sin 3 x cos 2 z da: = — - sin 2 x cos 3 x — — ; cos 3 a; = - cos 5 a; cos 3 a;. 
5 15 5 3 

This may also be integrated in the following way: 

f sin 3 x cos 2 x dx — f (1 — cos 2 a;) cos 2 a: sin a; da: 

(cos 4 x — cos 2 a;) d cos x — - cos 5 a; cos 3 x. 

5 3 

135. Exercises. In the following exercises use a table of 

integrals, and test results by differentiation: 



■S 



xdx 

(l - x y 

x 2 dx 
(2 + a;) 2 

a C dx 

5 -J 



6 -J 



V3a: 2 -7 
a; 2 (4 - a: 2 ) 



V; 



dz. 



7 -J ; 

8. f\/3 -4a: 2 da:. 

9 C ^_ . 

' J (1 + 2/ 2 ) 2 



wdw 



°* J "(1 + O 2 
1. JV (1 - x*)idx. 
dx 






V3" 



2z-4 



dx 



xVSx 2 -2x-\-4: 

4. fV4 -3u 2 v?du. 

5. fsin 4 0cos 2 0d0. 

6. Csec z ddd. 

7. f tan 5 d0. 

8. fVW^lcxidx. 



Sometimes it is advantageous to make a slight transformation of the 
function before applying the formula. For example, exercise 5 above 
may be integrated by formula 120, but if we write it in the form 

/ dx 1_ /» dx 

V3 a; 2 - 7 ~ VsJ Va^f' 
it can be integrated more easily by formula 92. 



§§136-137 SPECIAL METHODS OF INTEGRATION 193 

19. Integrate exercise 5 by the method suggested. 

20. Integrate exercise 8 by a similar method. 

136. Some Special Methods of Integration. There are sev- 
eral devices, more or less general in their application, for reducing 
a function to an integrable form, that is, for bringing it to a form 
in which its integral is known, or in which it can be integrated by 
a table. We shall present only a few of these devices. They are 
the following: 

Integration by algebraic substitution; 

Integration by trigonometric substitution; 

Integration by parts. 
Sometimes it is necessary to apply one of these methods before 
using the table of integrals, and sometimes a function which can 
be integrated by the table can be integrated more readily by one 
of these methods. 

137. Integration by Algebraic Substitution. When an alge- 
braic function of x contains no irrationality except fractional 
powers of x, it may be rationalized, and so brought to an inte- 
grable form by the substitution 

i_ 
(a) x m = z, 

where m is the least common multiple of the denominators of the 
fractional powers of x. 

Example 1. C dx . 

Let x* = z; then x = z 6 , dx = Qz b dz, x* = z 3 , x* = z 2 , 

and /ra= 6 /iS= 6 /S= 6 /(^ 2+1+ ^i)^ 

= 2z 3 + 3z 2 + 6z + 61og(2- 1), 
= 2 x* + 3 rc3 + 6 x* + 6 log (x* - 1). 

When an algebraic function of x contains no irrationality except 
fractional powers of a + bx, it may be rationalized and so made 
integrable by the substitution 

l 
08) (a + bx) m = z, 



194 INTEGRAL CALCULUS §138 

where m is the least common multiple of the denominators of the 
fractional powers of a + bx. 



Example 2. 



I 



ydy 



V(3 2/ + 2) 2 



Set (3y + 2)i=z; then y =\ (z 3 - 2), dy = z 2 dz, (3?/ + 2)1 = z 2 , 

o 

j r V dy 1 r (z 3 — 2)z 2 c?2 1/- M ON , 1 . 2 

and ( .. — = - I r = - I (z 3 — 2)dz = — z 4 — - z 

J^/(3|/ + 2) 2 3 J z 2 3 J y ) 12 3 

= F2 ( " 3 ~ 8) = 12 (3 y " 6) ^^7+2 = 4 {y - 2) ^^+"2. 

Example 3. f x • 

J a; + 2 + Va; + 2 

Set J" VaT+T = z; then a; = z 2 - 2, da; = 2 z dz, 

and f ^^ = 2 f^ = 2f-^ = 21og(z + l) 

= 21og(v / x + 2 + l). 

Example 4. J'V / x (Va7+ 1) dx. 

Set Va; = 2; then a; = z 2 , da; = 2 2 dz, 

and fV x (Vx + i) dx = 2 f Vz + 1 z 2 dz. 

Now this can be integrated by the table (formula 76), or we may pro- 
ceed as follows: 

Set Vz + 1 = u; then z = u 2 — 1, dz = 2 u du, 

and 2 f\/z + lz 2 dz = 4 f(it 2 - l) 2 u 2 du, 

= T fo (15 w 4 - 42 w 2 + 35) w 3 , 
= T o3 (15 z 2 - 12z + 8)(z + l)l 

/ V x (Vx+l)dx = ifo (15 x - 12 Vx + 8) (Vx + I)" 1 . 

138. Exercises. 

x f ^ <*c , 3. fVx + lxdx. 

^ xi + 1 

2 . r rf y . 4. r— -i? — . 

J l + Vy J Ve (o + 1) 



139 SPECIAL METHODS OF INTEGRATION 195 

6 r xdx . 19 r dy_ 






xdx 



9 -/ 



xdx Set aj = z 2 , and then use the table. 



V(x+1) 2 ,. r x 2 dx 



"■/: 






10 



■J: 



d?/ 



15. 



J (2 + z 2 )i ' 



2/ + 5- V2/ + 5 Set2+x 2 = * 2 . 

Besides the substitutions (a) and jj8) there are various others, 
such as the one suggested in exercise 15, but we shall discuss 
them no further. The " devices" of Arts. 124 and 127 are virtu- 
ally methods of substitution. 

139. Integration of cos m # sin n z. Formula 179 or 180 of the 

table may of course be used in this case, but when either m or n 
is a positive odd integer the following method is simpler. Suppose, 
first, that m is a positive odd integer. Then 

/r» m — l 

cos m x sin n xdx = 1(1 — sin 2 x) 2 sin n x d sin x. 

Since m is a positive odd integer, — = — is an integer, and conse- 

m — 1 

quently the binomial (1 — sin 2 x) 2 can be expanded and the last 
integrand can then be integrated term by term, whatever value 
n may have. 

So, too, if n is a positive odd integer, we have the identity 

I cos m x sin" xdx = — J (1 — cos 2 x) 2 cos m x d cos x, 

which can be integrated term by term, whatever may be the value 
of m. 

Example 1. 
f cos 3 xVsin x dx = f (1— sin 2 z) Vsin z d sin £ = f (Vsinz — sin^z)dsina; 
= | sin^ x — ? sin2 x = -fa (7 — 3 sin 2 x) sin^ x. 



196 INTEGRAL CALCULUS §§140-141 

The product or quotient of powers of any two trigonometric 
functions of x can be brought to the form cos m # sin n z, and the 
method of this article applies. 

Example 2. 

tan 3 xdx r sin 3 x r (l — cos 2 a) c? cos % 



/ tan d x dx _ ^ sm 6 x r 

Sfifi2 X J f»f>S2 X J 



3 

sec2 x J cos 2 x J cos 2 x 



— — ( (COS 2 x — COS2 x)d COS X = 2 COS 2 a; -f- i C osi X 

_ 2 3 + cos 2 % 
3 Vcos x 

140. Exercises. 

1. Csin z xdx. 4. ftan 5 0d0. 

2. fcos*xdx. 5. C-^Azdz- 
J J Vsecz 

3. J"cos 5 0d0. 6 . f s in 3 zVlJeczdz. 

141. Integration by Trigonometric Substitution. An alge- 
braic function which contains an irrational expression of the 
form Va 2 =t x 2 or vx 2 — a 2 can sometimes be rationalized and 
rendered integrable by a trigonometric substitution, as follows: 

a. When the irrationality Va 2 — x 2 occurs, 

set x = a sin 6, or a cos 0. 
|9. When the irrationality Va 2 + a: 2 occurs, 

set x = a tan 6, or a cot 6. 
y. When the irrationality Vx 2 — a 2 occurs, 

set x = a sec 0, or a esc 0. 

Example 1. f\/a 2 — x 2 cfo. 



J 



Set x = a sin 0; then Va 2 - z 2 = a cos d, dx = a cos d0, and 



x 2 cfo 


= a 2 ( cos 2 

= - , cos e 
a 


Odd = 


a 2 , . 
= 2 (* 


sine 


2 -x 2 
a 



Now sin e = - , cos — , and = sin -1 - 

a a a 

Therefore, CVa 2 - x 2 dx = $ \x Va 2 -x 2 + a 2 sin" 1 -1 • 



§142 SPECIAL METHODS OF INTEGRATION 197 

This could also have been integrated by the substitution x = a cos 0. 

Example 2. f J! — • 

J V(3 + 2?/ 2 ) 3 

This may be written in the form — f ■ . 

2* J V(i + </ 2 ) 3 

Set t/ = Vitan0; then V(i + y 2 Y =f V|sec 3 0, cfc/ = VUetfedd, 

, /• d?/ /-Vf sec 2 ddd 2 r dd 2 r j 2 . 

and I = I — = - = - ( cos dd = - sin 0. 

%/ V(| + ?/ 2 ) 3 •/ f\/fsec 3 3 J sec 3 J 3 

at „ tan y 

Now sin = 



sec V| -f y 2 
dy 1 2 y = 1 y 



Therefore, f , y — = — . y = - , y 

JV(3 + 2y 2 y 2l 3VI+2/ 2 3 V3 + 2 y 2 



Example 3. f 



2 V(z 2 - a 2 ) 3 



Set z = a sec 0; then v (z 2 — a 2 ) 3 = a 3 tan 3 0, dz = a sec tan d0, and 
V(z 2 - a 2 ) 2 



/ dz r asec0tan0a0 _ 1 /- cos d , 
^2 y/(# — a 2)3 J a 2 sec 2 • a 3 tan 3 a 4 J sin 2 



Now cos = = - , and sin = 



~, (( -Al -l)dsme = -\(-±- +sin ^ . 
av \sin 2 / a 4 \sin0 / 

Vz 2 - a 2 



sec z z 

Vz 2 - a 2 



Hence f * = _ 1(_^ + ^i^- 2 ) = 1 ^ 2 - 2^ . 

^ z 2 V(z 2 - a 2 ) 3 a Vz 2 - a 2 « ' a 4 z Vz 2 - a 2 



142. Exercises. 



z 2 dz - /» rfu 

'rr^ ' J (w 2 + 

^ * fV3z 2 -2 



f ax 6 f 

J (l-4x 2 )t 

;3 Vx 2 — 1 



C?Z. 

;^2 S Z 

3 - /wfcf" 7 -/(al-*f)t<fa, 



4. f ^ ffirtf. Set aj = a cos 3 0, and then 

(x 2 — a 2 ) I integrate by the table. 



198 INTEGRAL CALCULUS §143 

143. Integration by Parts. From the formula for the differ- 
entiation of a product 

udv + vdu = d(uv), 
we get by integration 

I udv + / vdu = j d(uv) = uv, 

whence 

(a) I udv = uv — I v du. 

This is an exceedingly useful formula and is known as the formula 
for integration by parts, and the process of applying it is termed 
integrating by parts. 

The following examples show how to use this formula. 

Example 1. fsiii~ 1 xdx. 

Let u = sm~ l z, dv = dx; then du = — , v = x. 

Vl-x 2 

Substituting in formula (a), we have 

/sin -1 x dx = x sin -1 x — C - = x sin -1 x + Vl — x 2 . 

J -v/l _ ^2 



X' 



Example 2. Cx 2 \ogxdx. 



We are to regard a; 2 log x dx as of the form u dv. In this case there are 
several ways of choosing u and dv. Thus we may let 

u = x 2 log x, and dv ±= dx, 

or u = x 2 , and dv = log x dx, 

or u — x, and dv = x log x dx, 

or u = x log x, and dv = x dx, 

or u = log x, and dv = x 2 dx. 

The last choice will be found the most effective. 

dx x^ 

Since u = log x, and dv = x 2 dx, then du = — , and v = — • 

x 3 

Substituting in formula (a), we get 

\x 2 log x dx = i x z log z — \ Cx 2 dx 
= | x 3 log z — I x z = i 2 3 (3 log x — 1) 



§143 SPECIAL METHODS OF INTEGRATION 199 

Let us solve this by one of the other selections of u and dv. Let 
u = x 2 log x, dv = dx; 
then du = 2 x log x dx + x dx, v = x, and substituting in (a), 

J x 2 log x dx = x 3 log x — 2 f x 2 log xdx — \ x 2 dx, 
3 f x 2 log x dx = x 3 log x — \ x 3 = i x 3 (3 log x — 1), 
and J x 2 log x dx = | x 3 (3 log x — 1). 

Example 3. \ xe ax dx. 

/x x 

- e ax d(ax), and then let u = -> dy= e ax d(ax) ; 
a a 

then du = — , v = e ax , and 
a 



,m _az 



/xe ax cta = - — - — f — dx = = — - (ax — 1). 
a J a a a 2 a 2 

Example 4. Cx 2 e x dx. 

Two applications of formula (a) are necessary here. Let 
u — x 2 , dv = e x dx; then du = 2 x dx, and v = e x . 

Then Cx 2 e x dx = xV - 2 Cxe x dx. 

The last integral is simpler than the given one because it has a lower 
power of x. 

It may be integrated by a second application of formula (a). This is. 
unnecessary, however, because we need only to set a = 1 in example 3 

above, and we have Cxe x dx = e x (x — 1). Therefore 

Cx 2 e x dx = xV - 2 e x (x - 1) = e x (x 2 - 2 x + 2). 

In applying formula (a) there are usually several ways of select- 
ing u and dv. They should of course be so chosen that the inte- 
gral in the second member of the equation shall be simpler than 
the given integral, but usually this can be determined only by trial. 



200 INTEGRAL CALCULUS §144 

144. Exercises. Integrate the following by parts: 

1. fxcosxdx. 7. ftzn-ixdx. 13< f 2 VT=zdz. 

2. fx 2 smxdx. q ftfln-i-v/^^ -, , r . 

j a. j tan v x cte. 14- j^ sm f cos t dt 

3. fa: tan 2 x dx. 9. f^t s j n -i x dXm # 

^ J 15. j sm 2 cos 2 d0. 

4. fsin 2 xcte. 10. fVe-*^. . 
17 J 16. J xlogxdx. 



udu. 



5. fcos 2 zcfa. 11. f log «, 
J J 17. f-^' 

6. fx*e x dx. 12. JV sin zdz. ' V4 " 

Use any method to integrate the following: 

J 7x + 5x 7 ' " J x 2 - s / x ~ZTi 

19. I , on rcos 3 ede 

J Vl2-4:x-x* 30. I — — • 

17 VsinJ 

20 f rf:r 

^ -\A 2 — 4 a: + 12 31. fsin 5 cos 2 6 d 6. 

21. r Vr +^ d;;. 32. f /* ■ 

22 ' /u+fe' 33. /-f^=- 

^ V4 — 2z 

. (sin 3 x sin 2 x dx. . 

r du 34. j 1 dx. 

C udu . 35 - fxi^S+2xidx. 

' J 8 + u 3 

rl£<fc_. 36 . / Vl+2*-^ 

' J (l-2x) 2 J * 2 

V' /(Tr§jT ^ 37. JvS5=I?d* 

28 ' /Ssfe" 38 ' JV4^25^ 



CHAPTER XX 

SIMPLE APPLICATIONS OF INTEGRATION IN 
GEOMETRY 

145. Illustrative Problems. 

Problem 1. To determine the curve or system of curves whose slope 
is everywhere proportional to the abscissa of the point of contact. 
Solution. By hypothesis 

D x y = ax, 

where a is a known constant. Integrating both sides of this equation, 
we have 

| D x y dx = \ dy = fax dx + C, 
and J J J 

(a) y = h ax 2 + C, 

C being the constant of integration. The curve sought is seen to be a 
parabola whose axis is OY. Since C may take all values between — oo 
and +00, equation (a) represents not one parabola but an entire system 
of parabolas whose common axis is the y-axis and each curve of which 
satisfies the requirements of the problem. C is termed the parameter of 
this system of curves. 

By giving a suitable value to C, we may select a curve of the system to 
satisfy some other condition, as, for example, that it should pass through 
the point (2, 1). 

Substituting 2 and 1 for x and y in (a), we have 

1 = 2 a + C, whence C = 1 - 2 a. 
Then y=±ax 2 + l-2a 

is the particular curve of the system that passes through (2, 1). 

From the foregoing solution it is evident that every equation 
in x and y that arises from an integration contains the constant of 
integration as a parameter, and consequently represents a system 
of curves. 

201 



202 



INTEGRAL CALCULUS 



§145 



Problem 2. To determine the system of curves whose subtangent ia 
constant. 
Solution. The equation of the tangent to a curve is 

y — 2/1 = Dyi(x - Xi). 
The subtangent is the line AB of the 
rpf figure. If x be the abscissa of A, AB = 

Xi — x. When y = in the equation 
of the tangent, x = x ; therefore 
-2/i = Ztyi(z - xi), 




Xi — x = ~- = subtangent. 
Dyi 



and 
Now by the condition of the problem 

where a is a known constant. Therefore, omitting subscripts, 



aDy = y, ady = y dx, a 



dy 

y 



dx. 



Hence 

and 



./?-/*+*, 



2/ 

a log 7/ = x + C, 

C being the constant of integration. This is the equation of the system 
sought. It may be transformed as follows: 



log y = 



x + C 



x + C 



C x 

= e° e a , 



or, writing k for the constant e°, 

X 

y = ke a i 

and is now seen to be the equation of a system of exponential curves. 
The parameter is k. This is the complete solution of the problem. As 
in the preceding problem, a curve of the system may be found to fulfill 
some further condition. For example, if the curve is to pass through the 
point (0, 1), we have 

1 = ke° = k, 



and hence y = e a 

is that curve of the system which passes through the point (0, 1). 



§145 APPLICATIONS OF INTEGRATION IN GEOMETRY 203 



Problem 3. To determine the system of curves in which the perpen- 
dicular from the foot of the ordinate upon 
the tangent has a constant length. 

Solution. The equation of the tangent 
may be written 

y — Dy x • x + XiDyi — y x = 0. 

The coordinates of the foot of the ordi- 
nate are (x h 0), and consequently, if a be 
the length of the perpendicular in ques- 
tion, we have 



— Dyi • xi + XiDyi — y\ _ 
Vl + {Dy,) 2 



/ 

-yi 




vi + (D Vl y 



Dropping subscripts, we have 



VI 



1 + (Dy) 2 
D*y _ _l 

n2 a 



Vy 2 - a 2 
Then, from the last equation, 



t 

dy = 
Vy 2 - a 2 



dx 
a 



f 



dy 



dx 



Vy 2 



/ax . 
a 



C. 



Integrating this by formula 92 of the table of integrals, 

log (y + Vy^^a 2 ) = - + C. 
a 

This is the equation of the system sought. We shall be able to recognize 
the nature of the curves of this system, after solving the equation for y. 



y + v y 2 — a 2 = 
Vy' 



+ c 



2 — a 2 = —y + e a , 
y*-a 2 = y 2 -2ye a +e Va ', 



and finally 



r ? +c -(- + c\] 

y=ile a + a 2 e Va 'J, 

wherein it is seen that the curves that satisfy the conditions of the problem 



204 INTEGRAL CALCULUS §146 

belong to the catenary type. In particular let us consider that curve of 
the system for which C = log a. We have then e c = a, and 

y== 2\ ea /' 

which is the familiar catenary. See Art. 87, exercise 26. 

146. Exercises. Determine the following systems of curves: 

1. Whose slope is proportional to the ordinate of the point of contact. 

2. Whose slope is proportional to - of the point of contact. 

x 

3. Whose slope is proportional to - of the point of contact. 

y 

4. Whose slope is proportional to xy of the point of contact. 

5. Whose subtangent is proportional to x m+l of the point of contact. 

6. Whose subnormal is constant. 

7. In which the portion of the \ \\ included between the coordi- 
nate axes is bisected at the point of contact. 

8. In which the tangent, ordinate of point of contact, and OX make a 
triangle of constant area. 

In the following exercises use formula (a) of Art. 97, — ^— = tan <j>, 

dp 

and thus get the equation of the curve in polar coordinates. 

9. Determine the curve which cuts its radii vectores at a constant 
angle. 

10. Determine the curve in which the radius vector, the tangent, and 
the initial line form an isosceles triangle. (3 solutions.) 

11. Determine the curve in which <t> = 2 d. 

12. Determine the curve in which <£ = § 0. 



CHAPTER XXI 

APPLICATIONS OF INTEGRATION IN KINEMATICS 

147. To Determine the Motion of a Body when the Velocity 
is Given. It was shown in Chapter IX that the velocity of a 
moving body is equal to the time derivative of the distance trav- 
ersed, that is, 

(a) -r = v, or ds = v dt. 

In the problem before us the velocity, v, is supposed to be given, 
and may be a constant, or a function of t, or of s, or of both t and 
s, according to the nature of the motion. Integrating equation (a), 
we have 

(b) s=fvdt + C, 

where C is the constant of integration whose value will be deter- 
mined by the initial conditions of the problem. 

We shall apply this formula to the solution of a few simple 
problems of rectilinear motion. 

Problem 1. Near the earth's surface the velocity of a falling body 
acted upon by gravity alone is found to be given by the formula 

v = 32 1 feet per second, 

where t is the time in seconds since the fall began, and the resistance of 
the air is neglected. Determine the motion. 

ds 
Solution. We have given — • = 32 t. Integrating this, we have 

CLli 

fds= C32tdt + C, or s = 16* 2 + C. 

Let s = So when t = 0; then C = s , and the last equation becomes 

(1) s=16i 2 + s . 

Let A be the point where the fall begins (t = 0) and the point from 

205 



206 INTEGRAL CALCULUS §147 

which we measure s. Then OA = s . Of course it would be simpler to 

measure s from 0, so that s = when t = 0. In that case s 3 = 


and our formula becomes 

£=fl (2) s = 16* 2 . 

This equation and the given one, v = 32 t, tells us all about the 
motion of the body. Observe that we choose the positive direc- 
B tion of s and v to be downward. 

Problem 2. A body is thrown directly upward with a velocity 
of 160 feet per second. Neglecting the resistance of 
the air, determine the motion. 

Solution. The entire velocity is made up of the velocity 
due to gravity and the velocity due to the throw, the latter 
being contrary in direction to the former. Choosing the 
positive direction of s and v to be downward, we have q 



v = S = 32 t - 160. 



t=o 



t = s 



£=10 



Integrating this equation, we have 

C ds =JS2 t dt -J 160 dt +C, 

or s = 16 1 2 - 160 1 + C. 

For convenience we measure s from O, the point at which the body was 
thrown, so that s = when t = 0; then C — 0, and the equations that 
describe the motion are 

(3) (a) v = 32 t - 160 = 32 (t - 5), 

(b) s = 16 & - lfflt = 16 tit - 10). 

Equation (b) shows that s is — so long as t < 10, which means that 
for the first 10 seconds of its flight the body is above its starting point 
O. From (a) it is seen that v is — , 0, or +, according as t is < 5, = 5, or 
> 5, and this means that the body rises for the first 5 seconds of its flight, 
comes to rest at A when t = 5, and then falls. From (b) we see that s 
passes O again 10 seconds after the start. Writing 5 for t in (b), we get 
s = —400, which is the height in feet to which the body rises. Writing 
10 for t in (a), we get v = 160, which is the velocity in feet per second with 
which the body passes O in its fall. This is the same velocity as that 
with which it rose from O, a result that could have been foreseen. 

Observe that in each of these problems 

Acceleration = — = 32 feet per second. 



§147 APPLICATIONS OF INTEGRATION IN KINEMATICS 207 

Problem 3. Harmonic Motion. 

A body is moving in a straight line and its velocity at time t is given by 
the equation 
(a) v — — a sin fit, 

where a and n are known constants: determine the motion. 

ds 
Solution. We have -~ = — a sin nt. Integrating this, 
at 



/ds = | sin ^t did + C, or s = - cos nt + C. 



Let us choose the point from which s is measured so that s = - when 




t = 0. Then - = - cos + C, and C = 0. And the formula which ex- 

presses s in terms of t is 

(b) s = - cos /xJ. 

Formulae (a) and (b) completely describe the motion of the body. Let 
be the point from which s is measured, and let A be the body's position 

when t = 0; then OA = - , and this is plainly the body's greatest distance 

to the right of 0. Let A' be the 

body's position when t = - : 

then OA' = - ^ and A ; is the 

body's greatest distance to the 
left of 0. 

Let the student show from a study of equations (a) and (b) that the 

body, starting at A, moves to A' and back again to A in— - units of time; 

that it repeats this oscillation in every succeeding — units of time; that at 

each passage through it has its numerically greatest velocity of +a 
or —a; and that it passes through any point of its path in the same 

direction every — - units of time. 

The motion is termed harmonic. Such is the motion of swinging pen- 
dulums, vibrating tuning forks and violin strings, of weights suspended 
by elastic cords or coiled springs and vibrating up and down, etc. The 
length of A A' is termed the amplitude of the oscillation, and the time 

of making a complete oscillation is termed the period. The period is 

t\ 

— - units of time. 



208 INTEGRAL CALCULUS §148 



(c) t=- cos- 1 &■ 

v a 



To express t in terms of s, we solve (b) for t and get 

) 

Differentiating (a), we get 

Acceleration = — 
at 

whence it appears that the acceleration is proportional to the body's dis- 
tance from 0, the center of the path, and is always toward 0. 

148. Exercises. When necessary draw a figure to illustrate 
the problem. 

1. From an elevation 150 feet above the ground a body is thrown 
downward with a velocity of 20 feet per second. Show that it will strike 
the ground in 2\ seconds with a velocity of 100 feet per second. 

2. With what velocity would a body have to be thrown upward from 
an elevation 200 feet above the ground in order that it should remain 
in the air 10 seconds? How high would it rise and with what velocity 
would it reach the ground? 

3. A balloon 1000 feet above the earth is j descer \ dmg | with a speed of 

( ascending ) 

60 feet per second. A stone is dropped over the edge of the car. How 
long is it in the air, and what is its velocity when it strikes the ground? 
In the second case, what is the highest altitude the stone attains ? 

4. In the presence of certain acids, sugar is converted into other sub- 
stances. The rapidity of the conversion at time t is found to be pro- 
portional to the amount of sugar remaining unconverted. That is, if a 
be the amount of sugar at the start and x the amount converted in time t, 
then 

dx ,' , 

¥ _, _„(„-*), 

where n is a known constant. Determine the amount of sugar converted 
in time t. See Arts. 80 and 72. 

5. In certain chemical reactions it is found that if a is the amount of 
a substance at the start, and x the amount of this substance transformed 
in time t, then the rapidity of the reaction is given by the formula 

v = n(a — x) 2 . 

Find the amount transformed in time t. 

6. Solve exercise 5 on the supposition that v — ix{a — a:) 3 , and also 
for the general case v = n(a — x) n . 

7. A stone is dropped from the edge of a well into the water, and 



/ 



§149 APPLICATIONS OF INTEGRATION IN KINEMATICS 209 



3 1 seconds elapse before the splash is heard. If sound travels in air at 
the rate of 1090 feet per second, how deep is the well? 

In the following problems the motion is rectilinear. Determine the 
motion. 

8. v = a Vs. 10. v = a cos fit. 12. v = ae? H . 

9. v = V a 2 - n 2 s\ 11. v = n 2 s. 13. i; = a(l + sin*). 

149. To Determine the Motion of a Body when the Acceler- 
ation is Given. We learned in Chapter IX that the acceleration 
of a moving body is the first time derivative of the velocity, or 
the second time derivative of the distance traversed. That is, 
if a be the acceleration, 

d 2 s _ dv _ 

~dP~~dt~ a) 

where a may be a constant, or a function of any or all of the 
variables t, s, and v, according to the nature of the motion. By 
integrating this equation we get equations expressing two of the 
quantities s, v, t in terms of the third, and such equations consti- 
tute a solution of the problem. 

Problem 1. Motion of a Body Falling near the Earth's Surface. 

To a body free to fall near the earth's surface, gravity imparts a con- 
stant acceleration. Denoting this acceleration by g and neglecting the 
resistance of the air, we shall determine the motion. 

First Solution. 

S - |= 9, whence /g<tt = /<fo=/ 9 <a + C, 

and, since g is a constant, Q s = o 

ds 



(1) 



dt 



= v = gt + C. 



Integrating again we have 

jds = fvdt + C =fgtdt +CCdt + C", 
whence 

(2) 8=fvdt + C = lgt 2 + Ct + C. 

If when t = the distance fallen be denoted by s and 



t=o 



>s„ 



S=..S 

v=v 



210 INTEGRAL CALCULUS §149 

the velocity by v , then from (1) C = v , and from (2) C'= s , so that we 
have finally 

(3) v = gt + v and s = \ gt 2 + ^o^ + s . 

These equations completely describe the motion of a freely falling body. 
The value of g is found by experiment to be about 32 feet per second, 
whence it appears that equations (3), Art. 147, are special cases (v = —160, 
and So = 0) of equations (3) above. If we choose the origin for s so that 
s = when t = 0, and if the body has no initial velocity, then s = and 
v o = 0, and equations (3) take the form 

(3 ; ) v = gt, s = \ gt 2 , or v = 32 1, s = 16 1 2 . 

This is Problem 1 of Art. 147. 

Second Solution. 

We shall first derive a new derivative formula for the acceleration. 
We have 

dt 2 dt ds dt ds 

dv 
Hence for the problem before us we may write v — = g, and on integrat- 
es 

ing we have 

/r K 

vdv = i gds + —, and v 2 = 2 gs + K. 

Therefore, 

(4) | = v = V2gs + K. 



dt= ds 



From this equation follow 

_- and C*-±f*®M±m., 

whence 

(5) t = V2g S + K + K\ 

9 
Setting v = v and s = s when t = 0, we have, from (4), 

v = V2 gs + K, whence K = v 2 — 2 gs 0) 
and from (5), 

= a/2 gs + K + K', whence K' = -v . 

Substituting these values for K and K' in (4) and (5), we get 



(6) » = V2g(s-So)+vJ and t = V2 g(s ~ S °> + -*?■ 

5/ 



§149 APPLICATIONS OF INTEGRATION IN KINEMATICS 211 

The sole difference between solutions (3) and solutions (6) is that in (3) 
v and s are expressed in terms of t, while in (6) v and t are expressed in 
terms of s. Either set of solutions can be derived directly from the other 
set. Thus on eliminating the radical from the equations of (6) we get 
the first equation of (3), and on squaring the last equation of (6) and 
solving for s we get the last equation of (3) . 

If we suppose s = and v = when t = 0, then s = and v = 0, and 
equations (6) become 

(6') • v = V2is and t = ^, ■ 

results which can be obtained directly from (3')- ' 

Problem 2. Harmonic Motion. 

The resultant of the forces acting upon a body is such that the body 
moves along a straight line, toward a fixed point of that line, with an 
acceleration proportional to its distance, s, from 0. Determine the mo- 
tion. 

Solution. Since the acceleration is directed toward 0, it will be + 
when s is — , and — when s is +. We have then 

d 2 s 9 dv 9 

_=_ A or . s — *. 

Multiplying the last equation by ds and integrating, there results 

and therefore 

(1) » = "J = - Vc - M 2 s 2 . 

We write the negative sign with the radical, although either sign may be 
used. Writing this equation in the form 

dt=- ds 



Vc - fi 2 S 2 

and integrating, we get, after certain simple transformations, 

(2) s = — cosU + c'), 

Substituting this value for s in (1), the result is 

(3) v = - V~c sin (nt + c'), 

a result which may also be obtained by differentiating (2). Suppose now 
that v = when t = 0; then = — Vc sin c', whence c' — 0, ± ir, 



212 INTEGRAL CALCULUS §150 

± 2 7T, . . . . Choosing c' = and writing a 2 for c to avoid the radical, 
we have finally 

(4) v = — a sin yi and s = - cos fd, 

which are identical with formulae (a) and (b) of problem 3, Art. 147. 
The motion is therefore harmonic, the body oscillating about 0. /x 2 is 
the acceleration at unit's distance from 0. 

It is a principle of physics that a solid homogeneous sphere 
exerts upon each particle of matter within it an attractive force 
such that, were the particle free to move, it would approach the 
center of the sphere with an acceleration proportional to its dis- 
tance from the center. If we assume the earth to be such a sphere 
and if we neglect the resistance of the air, then a body dropped 
down a straight shaft through the center of the earth would have 
the harmonic motion just described. It would oscillate back- 
ward and forward from surface to surface. To determine com- 
pletely the motion in such a case, we must know the constants 
a and y.. We know that at the earth's surface the acceleration 
is 32 feet per second. Now s is measured from the center of the 
earth, and when s = R = the earth's radius, 



32 = »*R, vl = y/§ , 



where R must be expressed in feet. Now when t = 0, s = R, 
and the second equation of (4), problem 2, gives 

a = nR = VS2R. 

Ijl and a are now known numerically and the problem can be solved. 

150. Exercises. 

1. In the problem of the body falling down a shaft through the center 
of the earth, find the velocity with which it would pass the center, .and 
the time required to reach the other side. 

2. What would be the result of giving c r , in equations (2) and (3) of 
problem 2 of the preceding article, the values ± w, ± 2 tt, . . . ? 

What would be the result of giving the sign + to the radical in equa- 
tion (1) of the same problem? 

3. Let the body of exercise 1 be brought to rest halfway between the 



§151 APPLICATIONS OF INTEGRATION IN KINEMATICS 213 



surface and the center, and then allowed to fall. Determine its motion 
completely. 

4. The diameter of the moon is 2160 miles, and at the surface gravity 
produces an acceleration of 5.3 feet per second. If a body fall down a 
straight shaft through the center of the moon, how long would it take it 
to go from surface to surface, and with what velocity would it pass the 
center? 

5. The conditions of the motion being the same as in problem 2 of 
Art. 147, except that the acceleration is away from instead of toward 
it, determine the motion. 

151. The Motion of a Body Falling to the Earth from a Great 
Distance above the Surface. The attraction of gravitation 
causes the body to move along the straight line joining it to the 
earth's center with an acceleration toward that 
center that is inversely proportional to the square s n v=o 

of the distance from the center. The effect of 
the earth's motion is not taken into account. 

Let Po be the body's position at the start 
(v = 0, t = 0), and P its position at time t. 
Let CP = so and CP = s, and denote by a 
the acceleration at P. The acceleration at Q 
we know to be g. Therefore, by the conditions 
of the problem, 

gR 2 



a'.gw 


1 . 1 

s 2 'R 2 ' 


whence a 


We have then 
(1) 


d 2 s 
dt 2 ~ 


dv 
ds 


gR 2 

s 2 




the negative sign being used because the direction of a is contrary 
to that of s. Integrating (1), we have 

/ vdv = —gR 2 f -g and v 2 = — — 



+ C 



Since s = s = CP when v = 0, C = — 



2gR'< 



Sn 



and 



(2) 



v =-£V2 



V S So' 



214 INTEGRAL CALCULUS §151 

which is the velocity acquired by a body falling toward the earth 
from a position distant s to a position distant s from the center 
C. (Why is the negative sign used with the radical in (2) ?) If we 

11 h 

set s = R and s = R + h, then = n/n , 7X . and 

s so R(R + h) 



(3) .--VS^-ij, 

which is the velocity acquired by a body falling to the earth's 
surface from a distance h above the surface. When h is very 

r> 

small compared with R, we have D , = 1 nearly, and then 

(4) v = -V2Jh, 

which is the law for a body falling near the surface of the earth, 
as we found in Art. 149, equation (6'). Equation (4) is not abso- 
lutely true, but, when h is only a few feet or a few hundred feet, R 

r> 

being 4000 miles, the fraction g , is so nearly 1 that the error 

made by the use of (4) instead of (3) is very small indeed. 
If in (2) we set s = R and s = oo , we have 

(5) v = ~V2^R, 

which is the velocity a body would acquire in falling to the earth 
from an infinite distance. Consequently if a body were projected 
vertically from the earth with this or a greater velocity, it would 
never return. Throughout the foregoing discussion the resistance 
of the air is neglected. 

Let the student obtain (5) by setting h = go in (3). 

To get an expression for the time, equation (2) must be inte- 
grated. We first put that equation into the form 



m 



ds D . /2 a . /so — s 



whence 



«=-*vf/*+« 



Vs e 



§§ 152-3 APPLICATIONS OF INTEGRATION IN KINEMATICS 215 

To integrate the first member, we note that 

C Vsds = so C ds _ 1 r js -2s)ds _ 

J Vs — s 2 J Vs s — s 2 2 J Vs s — s 2 

Now / , = cos _1 ( 1 -J, 

J Vsqs - s 2 V so / 

, 1 C(s — 2 s)ds r 

and o / / = V^ 

Therefore 



s s —s*. 

SqS 



•^(l-ty-Vv-f—Byftt + C, 



Since s = s when t = 0, we have 

Icos-U-l)^ and C = ^. 
Therefore, finally, 
(6) ^f^V^^ + lf.-cos-^l-^)]. 

152. Exercises. 

1. Show that the vertical velocity with which a projectile would have 
to be fired from the earth in order that it should never return is 6.96 miles 
per second. 

2. Find the percentage of error introduced by using formula (4) instead 
of formula (3) when h has the value (a) 100 miles, (b) 1 mile, (c) 100 feet. 

3. With what velocity and in what time would a body reach the earth 
falling from a distance R above the surface? from a distance 100,000 miles 
above the surface? 

4. What is the vertical velocity that a projectile fired from the surface 
of the moon would have to have in order that it should never return? 
Assume the moon's radius to be 1080 miles, and gravity on its surface 
to be 5.3 feet per second. 

5. With what velocity and in what time would a body reach the moon 
falling from a distance of 10,800 miles above the surface, the attraction 
of the earth being neglected? 

153. The Motion of a Projectile. 

Problem 1. A particle is projected from a fixed point O on the earth's 
surface, with a velocity v , in a direction that makes an angle 4> with the 
horizontal. Determine the motion, neglecting the resistance of the air. 




216 INTEGRAL CALCULUS §153 

Solution. The particle will of course remain in the same vertical plane 
throughout its flight; it will rise for a time, and will finally be brought 
back to the earth by the action of gravitation. Let be the origin and 

let the axes be the horizontal and ver- 
tical lines through that He in the 
vertical plane. Let P, with the co- 
ordinates x and y, be the particle's 
position t units of time after the be- 
ginning of the flight. If gravitation 
did not act, the body would move in 
a straight line with the constant ve- 
locity v , and the horizontal and ver- 
tical components of the velocity at any time t would have respectively the 
constant values v cos </> and v Q sin 4>. But gravitation engenders a velocity 
component downward of gt. Hence the velocity components of the par- 
ticle at P are 

ri\ dx dy • , . 

(1) — = v cos <f>, -f- = v sin <t> - gt. 
at at 

Integrating these equations, we find that the constants of integration are 
both when t = 0, and therefore 

(2) x = v cos • t y = v sin <j> • t — \ gt 2 . 

These are the parametric equations of the path of the projectile. From 
the last of these equations we see that y = 0, not only at the start when 

t = 0, but also when t = VoSm< t> which is the moment the projectile 

9 
strikes the earth. The distance of this point (A in the figure) from is 
termed the range of the projectile. This range is found by substituting 
the foregoing value of t in the first of equations (2) . This gives 

/on -d 2 v 2 sin $ cos tf> v 2 ■ , , „ t 2 v sin <j> 

(3) Range = =-sm2 <f>, when t = 

Q 9 9 

This range will plainly have its greatest value (for a given value of v ) 



when 4> = - . Hence 



*V i „_ ._ *„„A * _ V^Vo 



V2 



(3') Maximum range = ■ — , when <f>= - and t = 

To determine the direction in which the particle must be thrown with 
a given velocity v to strike a point A at a distance a from 0, that is, to 
have a range a, we have the equations 

(4) a = — sin 2 <j>, whence 2 <t> = sin -1 — n • 

9 v 2 



§153 APPLICATIONS OF INTEGRATION IN KINEMATICS 217 

Now when v 2 > ag there are two positive angles less than ir, each of which 

has the sine, — . If we denote one of these angles by 2 <f> h then the other is 

v — 2 £i. Hence there are two directions in which the particle may be 
thrown so as to strike A, these 
directions making with the hori- 

0i. 



zontal the angles 0i and 

Now if we set <fn = 7 — 
4 

+ a. That is, the two 



2 

a, then 




= - 0i - - 
2*4 

directions are equally inclined to 

the direction of maximum range. 

When Vq 2 = ag, sin 2 = 1, = j » 

which gives the maximum range. In that ,case a = 0, which means that 
the two paths coincide at the maximum range. 

When Vq 2 < ag, sin 2 <j> > 1 and 2 is imaginary. In that case the 
point A cannot be reached with the initial velocity v . 

To find the greatest height to which a particle rises for a given v and 
0, we have only to find the maximum value of y in (2). It is readily 

found that y is maximum when t = — , and on substituting this 

Q 
value for t in the equations of (2), we get the coordinates of the highest 
point of the flight. Hence for the point of greatest altitude, 0' , we have 



(5) 



t = 



Vq Sin 

9 



v 2 sin cos Vo 2 . , _ 
- = -sm20, y 



Q 



2<7 



snr0. 





Y 


' 






Y 


Directrix D 




/ F 


0' 


X' 






p 













\ x 













Since t is half the range, we see that 
the highest point is the mid-point of 
the path. 

The x and y equation of the path 
is obtained by eliminating t from 
the equations of (2) : it is 



(6) y = tan • x — 
From this we get 



2 vo 2 cos 2 



x\ 



(-£*»♦/- 



Vo SUT 



2g 



*■ 



218 



INTEGRAL CALCULUS 



§153 



Transforming to the point (- 1 sin2 0, - 2 — — -] for a new origin with 

\Ao ^9 I 

new axes parallel to the old, the equation becomes 

, 2 _ _ 2 v<? cos 2 <j> , 



(6') 



9 



The path of the projectile is thus seen to be a parabola turned downward, 
having its vertex at 0' , the highest point reached by the projectile. The 
equations of the directrix referred to the new axes and to the old axes 
are respectively 



(7) 



v = 



Vo z COS^ <f> 

29 



and y = 



zv cos^ 4> . v z sin^ <f> 



2g 



+ 



29 



By the last equation it is seen that the position of the directrix is inde- 
pendent of <f>. Therefore, the parabolic paths of all particles projected 
from in the same vertical plane and with the same velocity v have the 
same directrix, whatever be the direction in which they are thrown. It 
is to be noted that this common directrix is at a distance above equal 
to the height to which the body would rise if thrown vertically upward 
with a velocity v . 

N.B. If the projectile has a high velocity, the resistance of the air 
so modifies its motion that the foregoing formulae give very inaccurate 
results. The law governing the resistance of the air to the motion of 
a projectile is not accurately known. 

Problem 2. To determine the range on an inclined plane. 
Let OP be a plane through perpendicular to the plane of the pro- 
jectile's path and making an angle a 
with the horizontal plane. We seek to 
determine the point P where the pro- 
jectile pierces this plane. 

We have already found the equation 
of the path in the form 




tan <t> • x — 



1_ 



2 v 2 cos 2 <t> 



x\ 



The equation of OP is 

y = tan a • x. 

Solving these equations for x and y, we find, after easy reductions, the 

coordinates of P to be 



2 y 2 • / x 2v 2 

cos 0sm (0 — a),y= — — cos <f> sin 



gcosa 



gcos 



a) tan a. 



§153 APPLICATIONS OF INTEGRATION IN KINEMATICS 219 



Substituting these in the equation R = x cos a + y sin a, we have 



(8) 



Range on OP = R = 



2 y„ 2 



cos <f> sin (<f> — a). 



gcos'a 

To determine the value of <f> which makes R a maximum, we set 
u = cos <£ sin (</> — a) ; 
D^w = cos <f> cos (</> — «) — sin sin (<£ — «) = cos (2 <t> — a). 



2' 



7T ■ Q! 

4 + 2' 



then 

When D+u = 0, 2 - « 

Then n = cos (| + |) sin(^ - |) = sin 2 (| - |) 

= i[ 1 - C0S (l"° £ )] = J (1 " Sina) - 

Therefore 

(9) Maximum range on OP = -^ : — — = 



Itf 



g COS' a 



g(l + sin a) 



Setting = - + a in this equation and p for the maximum range, we have 



(10) 



_2_ 



COS0 




X 



Regarding p and as variables, this is the polar equation of a parabola 
turned downwards and having its focus at the pole 0. This parabola 
is the locus of points of maxi- 
mum range on all planes 
through (a variable) attained 
by projectiles fired from with 
the same velocity v . It can 
be proved that the parabolic 
path of each such projectile is 
tangent to the parabola (10). 
By revolving this parabola about its vertical axis, we get a surface which 
is the locus of all points of maximum range reached by projectiles fired 
from in all directions with the velocity v . This surface is termed a 
paraboloid of revolution. A projectile fired from with velocity v and 
in any direction would just reach this surface. All points inside the 
surface can be hit by such a projectile and no point outside it. 



220 INTEGRAL CALCULUS §154 

154. Exercises. 

1. The maximum range of a certain golf ball is 450 feet. What is 
the velocity of the drive, how high does the ball rise, and how long is it 
in the air? 

2. A cannon is sighted to make an angle of 30 degrees with the hori- 
zontal plane. What must be its muzzle velocity in order that it shall 
carry a mile? How high does the projectile rise, and how long is it in 
the air? 

3. A cannon has a maximum range of 10 miles. What is the muzzle 
velocity of the projectile, what is the greatest height it reaches, and in 
what time will it strike the ground? 

4. What initial velocity must a rifle ball have in order that it may 
rise only 1\ inches in a range of 500 feet? What is the time of flight? 

5. What must be the muzzle velocity of a cannon ball in order that 
it shall rise only 200 feet in a range of 10 miles? Find the angle of eleva- 
tion of the gun, and the time of flight. 

6. A boy can throw a stone 200 feet on level ground. How far can 
he throw it if he stands on a flat-car moving on a level track at a speed 
of 24 miles per hour, if he throws the stone : (a) in the direction the train 
is moving? (b) in the direction opposite to that of the train? (Leave out 
of account the height of the car above the ground.) 

7. A certain revolver has a muzzle velocity of 200 feet per second. 
What are the angles of elevation that will enable one with this revolver 
to hit a target 400 feet away on a level with the muzzle? 

8. A boy can throw a tennis ball 128 feet on a level (the distance of 
his hand from the ground being neglected). How far can he throw it in 
a cage 25 feet high, without hitting the ceiling? What is the most dis- 
tant point of the ceiling that he can hit with the ball? 

9. From an elevation h above the horizontal plane a projectile is 
hurled with a horizontal velocity of v . Determine the motion completely. 
Put cf> = in the formulae of Art. 153. 

10. Show that the tangential velocity of a projectile at any point of 
its path is equal to the velocity it would have acquired in falling, under 
the influence of gravitation alone, from the directrix to the point in 
question. 

11. Show that the foci of the paths of projectiles launched from the 
same point O with the same velocity y , but in different directions, lie on 
the surface of a sphere having O for center. 

12. Show that the foci of the paths of projectiles launched from~0 at 
the same angle with the horizontal plane, but with different velocities, lie 
on a cone whose vertex is at O. 



§ 155 APPLICATIONS OF INTEGRATION IN KINEMATICS 221 

13. A shell explodes in the air in such a way that the fragments all have 
the same velocity. Show that at any instant before they reach the 
ground these fragments lie on the surface of a sphere. 

14. From an elevation h above the horizontal plane a projectile is 
launched with a velocity of v and at an angle of elevation of <$>. If B 
be the foot of the perpendicular from the initial point to the ground (the 
horizontal plane), and if A be the point where the projectile strikes the 
ground, show that 

Vo cos 
9 



BA 



Vo sin <£ + VW sin 2 <f> + 2 gh j . 



15. A cannon in a fortress can be given an angle of elevation of 21° 6', 
and its muzzle velocity is 1200 feet per second. The muzzle of the gun 
is 108 feet above the water. How far from the base of the fortress will 
the projectile strike the water? 

155. Problem 3. 

A particle of mud is thrown from the rim of a carriage wheel when the 
carriage has a speed of m feet per second. Neglecting the resistance of 
the air, determine the motion of the particle of mud. 

Solution. Before being thrown off, the particle of mud is describing a 
cycloid, and as it leaves the rim its velocity is in the direction of the 
tangent to the cycloid at that point, P, and is equal in magnitude to 

ds ^ . , . dd 
dt' 



Vo = — = 2 a sin 
dt 



(Art. 104, exercise 4.) 



The slope of the tangent to the cycloid is -Or — 0) (Art. 93, problem 1). 



In the figure, angle APX = - (tt — 0). Since the carriage has a speed 

A 

of m feet per second, w r e have 



ad = mt, 



m 



dd 
dt 



Hence the tangential velocity at P is 
Vo = 2 7?z sin |0. The problem is now 
reduced to that of determining the mo- 
tion of a particle launched from the 
point P with a velocity 2 m sin^0, and 

at an angle of elevation of - Or — 0), being a constant. 




Hence we get 



the motion by setting v = 2 m sin 



and <t> = - (tt — 0) in the formulae of 



222 INTEGRAL CALCULUS §155 

Art. 153, problem 1. The axes of reference will be the horizontal and 
vertical lines through P. To find the highest point the particle of mud 
will reach, we make the foregoing substitutions in the last of equations 
(5) of Art. 153. The result is 

- _ Vo* sin 2 4> _ 4 m 2 sin 2 \ cos 2 \ _ m 2 sin 2 
V " 2g ~ " 2g 2g ' 

This is the particle's greatest height above PX. The greatest height 
from the ground is 

, m 2 sin 2 . M . 

h = — h a(l — cos 0). 

2 <7 

It is found without difficulty that h attains its maximum value when 

cos = — -?. and = cos _1 ( - n V This gives as the maximum value 

m 2 V m 1 

oih 

maximum h = ^— f 1 - • 

2gm? 

When m 2 < ag, this maximum altitude is never attained. 

The parametric equations of the path of the particle are obtained by 

substituting v = 2 m sin \ 0, and tf> = - (*■ — 0) in equations (2) of Art. 153, 

z 

and are 

x = 2msin 2 ^0-^, y = 2msin£0cos£0»£ — %gt 2 , 

which reduce to 

x = m [1 + cos (tt — 0)] t, y = m sin (tt — 0) • t — \ gt 2 . 

These last equations show that the motion may be conceived as com- 
posed of two motions: 

First, a translation in the positive direction of OX with a velocity m. 

Second, a projection from P 
with a velocity m in the direction 
of the tangent to the wheel at P. 
These results can also be ar- 
rived at in the following way: 

If the wheel did not rotate but 
merely slipped along the road 
with a speed m, the particle at P 
would have simply a horizontal 
O motion with a speed m. If, on 

the other hand, the wheel were to turn on its axle with a (tangential) 
velocity m, while the axle itself remained stationary, the particle ~ at P 
would have a velocity m along the tangent to the wheel at P. And the 
actual motion is obviously compounded of these two motions. 




§156 APPLICATIONS OF INTEGRATION IN KINEMATICS 223 

156. Exercises. 

1. If the wheels of a bicycle are 3 feet in diameter, and are without mud 
guards, and if the wheel is driven at a speed of 15 miles per hour, what is 
the highest flight of mud from the tires, and at what angle must it be 
thrown off to reach this height? 

2. A body has an initial velocity v and is subjected to a resistance 
that at any instant is proportional to its velocity at that instant. The 
equation of motion is then 

d 2 s dv ~, 

w> = Jt = - Kv - 

Determine the motion completely. This is approximately the law of re- 
sistance offered by air or water to a body whose velocity does not exceed 
10 or 12 feet per second. 

3. To a falling body, such as a raindrop, the air offers a resistance 
that at any instant is proportional to the square of the velocity at that 
instant. The equation of motion is therefore 

d 2 s dv „„ 2 

Determine the motion completely and show that there is a limiting 
velocity. 



CHAPTER XXII 

THE DEFINITE INTEGRAL. AREAS AND LENGTHS OF 

CURVES 

157. The Definite Integral. A definite integral of f(x) is the 
difference between two values of the integral of f(x), for two 
values of the argument x. 

The two values of the integral of f(x) for x = a and x = b may 
be denoted respectively by 

[//W^]^ C and [J /(x)dx l =6 + C ' 

the constant of integration, C, having of course the same value in 
each case. The difference between these values is represented by 



x 



f(x)dx, 

U a 

so that 

fmdx s r />(z)<fa] - r fmdx] , 

%) a |_t/ Jx = b \_<J Jx = a 

and it is this difference that we define to be the definite integral 
off(x) between the limits a and b, or, more briefly, the definite integral 
of f(x) from a to b. Because of their positions in the symbol 

/ f{x)dx, a is termed the lower limit and b the upper limit. 

I f(x)dx is termed the indefinite integral of f(x) to distinguish 

it from the definite integral I f(x)dx. It will be observed that the 

definite integral contains no arbitrary constant, which is the rea- 
son that it is called definite. Observe, further, that the definite 

integral, I f{x)dx, is not a function of x, but of the limits of integra- 
tion, a and b, 

224 



158159 



THE DEFINITE INTEGRAL 



225 



Examples. 

1. 



it 
2 f sin 2 x da; = - \x — sin x cos zl — 77 \x — sin x cos x] = 7- 

3. C -^r = [log (a - l)] 3 = log 2 - log 1 = log 2. 

^2 a; - 1 L J 2 

As in the last example, we shall write [^(z)]* for [Ffa)] ^[Ffa)] a > 

158. Exercises. Find the values of the following definite 
integrals : 

6. fcoaxdx. 11 1 ^(Sx^-x^dx. 

IT 

7 Cz^dz. 12 - Ptanfldfl. 

IT 

8. C (x*-2x z +x-2)dx. 13. Ptanflcte. 



1. 


) £ 3 efo. 


2. 


1 a; 3 da;. 
*2 


3. 


) x 3 dx. 


4. 


1 y 2 dy. 



*£ 



du 



5. J cosxdJa;. 10. J 



j Vi - u 2 

idy__ 



14. f 2 sin 2 0cos 2 0d0. 

IT 

15. J sin 4 cos 2 dd. 



- Vl — 2/ - 

159. The Derivative of Area: Cartesian Coordinates. Let U 
be the area aAPx, bounded by 
the curve y = f(x), the axis of 
x, and the ordinates of the 
points whose abscissas are a 
and x* This area is termed 
the area under the curve. We 
assume that fix) is real, single- 
valued, and continuous, so 
that the curve consists of a 
single, real, unbroken branch 

* The student ought to have no difficulty with this double use of a and x, 
at one time to denote points on the z-axis, and at another time to denote the 
abscissas of these points. 



■vr M 


Pi 


T 


P 




Ay 

N 




■ 






/ y 

u 


n 

■ 




'0 






fl 


X 




t 


I 


5 Ax I 


I 



226 INTEGRAL CALCULUS §160 

within an interval that contains the figure. The ordinate aA is 

any convenient starting place from which to measure U. Holding 

A and its ordinate aA fixed, let P move along the curve, carrying 

with it its ordinate y, or xP. The area U varies with x, the 

abscissa of P; that is, U is a function of x, and consequently has 

a derivative as to a;. We seek to determine this D X U. Let x take 

an increment Ax. The corresponding increment AU of U is the 

curvilinear quadrilateral PP'Rx. From the figure it is evident 

that 

MP'Rx > PP'Rx > PNRx. 

(In case PP' has a negative slope instead of a positive one, as in 
the figure, these and the following inequalities will be reversed.) 

Or {y + Ay) Ax > AU > yAx, 

Au 
whence y + Ay > -^- > y. 

Passing to limits, we have 

(1) D x u = y=f(x). 

160. The Area under a Curve Expressed as a Definite Inte- 
gral. From formula (1) of the preceding article we get by inte- 
gration 

U = fydx + C, 

y being a function of x, and C being the constant of integration. 
C depends upon the initial point A, for, since a is the abscissa of 
A (figure of Art. 159), it is plain that U = when x = a. Hence 



Therefore 



0=rA/daT| +C and C = -I Cy dx\ . 

U-fydx - \Jvte]z[fv*>] - x [f yd i= a ' 

From this equation it is evident that U is the definite integral of 
y (determined as a function of x from the equation of the curve), 
taken between the limits a and x, and we may write 

(2) U = f X ydx= ff(x)dx, 



§160 



AREAS 



227 



which is the formula for the area under the curve y = f{x), that is, 
for the area bounded by the curve, the x-axis, and the ordinates of the 
points whose abscissas are a and x. 

If we write the abscissa of the second curve point b instead of x 
to distinguish it from the argu- y 
ment of y, we have 



(3) 



U= fydx= f 

%J a *J a 



f{x)dx 



as the formula for the area 
shown in the figure. 

When the area to be deter- 
mined lies below the a>axis, like 
B in the next figure, y is nega- 
tive, and the definite integral, 

Jy dx, and consequently the 
b 




■b' o 





area B, will have a negative value.* 
If in the case shown in the figure we 
take the value of the definite integral, 



dx we shall have 



ydx=A -B + C,notA + B+C. 



f 

U a 

f 

%J a 

The process of finding the area under 
a curve is termed quadrature.^ 

Example 1. To find the area bounded 
by the cubical parabola y = x*, the z-axis, 
and the ordinates whose abscissas are a 
and b. 



* On a later page it will be shown why this definite integral is negative when 
y is negative. 

f The term quadrature is also made to include any integration without 
reference to the geometrical interpretation of the process. Thus, the integra- 
tion of J f(x) dx is termed a quadrature. 



228 



INTEGRAL CALCULUS 



160 



By formula (3) we have 



U = I ydx = ( x 3 dx = PJM 

J a J a L4Jo 



If a = 0, we have area OAS6 = 



*7 = 

b 4 



■jH-v b' 4 



C~° r~x?~\ ~ ° 

Again, U ' = | x? dx =\- \ 

J -a' L4J- a ' * 

and this area is negative because b' 4 < a' 4 . 

Example 2. To find the area bounded by the upper branch of the 
parabola y 2 = 2 mx, the z-axis, and the ordinate whose abscissa is b. 

In this case the lower limit is 0. By 
formula (3) we have 

U = £ y dx = V2mf o a* dx 

9 k a/9^ 




3 *- _„ 

2 6 V2m6 



If c be the ordinate of B, then c = v2 mb, and therefore t/ = f 6c. Now 
6c = area of rectangle OMBC. Therefore U = f OMBC. From the 
symmetry of the figure it follows that 

Parabolic segment SOS 7 =%BMM'B\ 

or, the area of the segment of the parabola cut off by a double ordinate 
is two-thirds the area of the circumscribing rectangle. 

Example 3. Let us find the area of the ellipse 

t. _l £. = 1 
a 2 ^ 6 2 

It will be simpler to find first the area of one quadrant. 

In this case y = -V a 2 — x 2 . Then by formula (3) 

U= C a ydx = - fV-z 2 )*' 
J a J n 

b 



dx 



b r /- j , 2 . .an- 

2 a L a J 



= — sm 1 1 = 
2 4 



ab 



/- 



§160 



AREAS 



229 



(We might of course integrate from 
—a to -\-a and so get half the area 
of the E.) 
Or we may proceed as follows: 

Let z = asin0; 

then Va 2 — x 2 = a cos 0, 

dx = a cos dd. 
As for the limits of integration, 

when x = 0, = 0, 

and when x = a, sin = l t = - • 



Consequently 
6 




C/ = - fVa 2 -z 2 efc = a& f 2 cos 2 d0 = ^[0 + sin0 cos 0l 



Hence 



a& 



7ra6 



The entire area of the ellipse is therefore irab. 
The last integration shows that when we substitute a new variable 

IN PLACE OF THE ARGUMENT IN THE INTEGRAND OF A DEFINITE INTEGRAL, 

WE MUST TAKE CARE TO MAKE PROPER 
CHANGES IN THE LIMITS OF THE IN- 
TEGRAL. 

Example 4. Let us find the area 
included between the parabolas 

y 2 = 2mx and x 2 = 2 my. 

From the figure 

V = OMBNO = 0MB AO - ONBAO. 




U "f. 



2mx^dx 



^«Mr-&M 



1m 




The coordinates of and B, the intersections of the curves, are (0, 0) and 
(2 m, 2 m). Hence 

„2to X 2 2 /— 

- — dx = - 
-0 2m 3 

= 2(2 m) 2 _ (2m) 2 (2m) 2 = 4m_ 2 

3 3 ~ 3 3 

It appears also that ONBAO = U. 

By interchanging x and y in formula (3), we have 

f xdy 



(3') 



U 



230 



INTEGRAL CALCULUS 



161 



as the formula which gives the area bounded by the curve x = f(y), the 
axis of y and the abscissas of the points whose ordinates are a and b. 

Example 5. The curve in the figure 
is the cubical parabola y = x*. Then 
x = y% and by formula (3') 




U= f xdy=C y idy 

a a 



Or we might proceed thus: 
Since y = x 3 , dy = 3 x 2 dx\ 

as for the limits, 
when y = a, x = a*, 

and when y = b, x = b*. 
Hence 



Let the student devise another method of finding the value of U. 



161. Exercises. 

1. Find the area cut off from the semicubical parabola y 2 — x 3 by 
the chord x = 4. 

2. Use formula (3'), and then formula (3), to find the" area bounded 
by the semicubical parabola y 2 = x 3 , the y-axis, and the chord y = 8. 

3. Find the area of the arch formed by the curve 2 y = x 2 (3 — x) 
and the z-axis. 

4. Find the area of each arch of the curve 3 y = 9 x — x 3 (Art. 65, 
exercise 1). 

2x 

5. Find the area bounded by the curve y = ~— — - , there-axis, and the 

1 + x 2 

ordinate of a flex. 

6. Find the area of each arch formed with the z-axis by the curve 

y 3 = x 3 (5 — x 2 ). 



Find also the area bounded by the curve, the x-axis, and the ordinate 
whose abscissa is 3. 



162 



AREAS 



231 



/»3 

Calculate for this curve the definite integral J y dx, and give its geo- 

•'0 

metric meaning. 
Use formula (3') to find area of one arch. 

7. Find the area cut from the curve y = (x z -f-.l) (x — 2) by the 
x-axis. 

8. Find the area bounded by the axes of coordinates and the parabola 

x 2 -\-y 2 = a*. 

9. Find the area bounded by the axes of coordinates and the curve 

x% + y% = a 3 - 
Solve by formula (3) and by formula (3'). 

10. Find the area of the circle by integration. 

11. Find the area of each arch of the curves y = sinx and y = cos x, 
and find also the area inclosed by these curves between any two consecu- 
tive intersections. 

12. Show that the line y = x — 4 bisects the arch of the curve 
8 y = x 4 — 4 x 3 . 

13. The curves 4 y = 8 x 2 — x A and 4 y = 4 x z — x A form with each 
other and with the z-axis several closed regions. Find the areas of these 
regions. 

14. Find the area inclosed by the curve 

?/ 2 = 4(3z 2 + 2) 2 (l-z 2 ) 3 . 

15. Find the area inclosed by the curve 

y i = (aji + 1)2 (2 - x 2 ) 3 . (Art. 87, exercise 18.) 

16. Find the area of each loop of the curve 

4 y 2 = (^ _ 1)2 ( 4 _ x iy ( Art# 87) exercise 21.) 

162. Areas of Curves Given by Parametric Equations. 

Example 1. Let us find the area of the ellipse from its parametric 

Y 



equations 

x = a cos 0, y = b sin 
We have dx = — a sin d0, 



J,s=a 
y dx 



= — ( sin cos — 0J-, 



/0=O 
sin 2 d0 



2 

7ra6 




Therefore the total area of the ellipse = 4 U = irafr. 
Compare this with the second solution in example 3, Art. 160. 



232 



INTEGRAL CALCULUS 



162 



Example 2. Let us find the area of the curve of exercise 9, Art. 91. 
x =f(t-3), y = t*(t-S). 
We have dx = 3 t(t - 2)dt, 



and 



fydx = 3J(t - 2){t - 3)t*dt = 3j> - 5 * 5 + 6 t*)dt 
= 3 (f " ¥ + ¥) = ™' 5(30i2 " 175i+ 252) ' 



Now it can be easily ascertained that the loop is traced in the direction 
of the arrows as t varies from to 3. Hence 



U= C* 3 ^z = ^[>(3(H 2 -175£ + 252)T = 
Jt=o 70 L Jo 



243 
70 



(-3) =-10.414. 



In order to understand how the integration from t = to t = 3 gives 
the area of the closed loop (and in this case 
with a negative sign), suppose a point, P, to 
start at and to trace the loop as t varies 
from to 3. It can be readily shown that 
as t varies from to 2, P traces the branch a 
from to M, the point of minimum x, and 
that as t varies from 2 to 3, P traces the 
branch b from M back to 0. Therefore 




and 



ydx 
f 

J t-- 



OaMB 



dx=-ObMB. 



In the first integral both y and dx are — , and 
consequently the integral and the area are +, 
while in the second integral y is — but dx is+, 
and both the integral and the area are — .* 
Hence 



2 y dx + f t=3 y dx = OaMB - ObMB 
= —area of loop. 
These results may be verified by calculation of the integrals involved. 

* On a later page this will be explained more fully. 



f ydx 



f. 



/ 



§ 163 AREAS 233 

Thus 



f' ==2 ydx = Mt*(ZOP - 175 t + 252)T 

= |? (22)= 10.057 ==OaM£. 

f' =3 ?/ dx =Mt 5 (30P - 175 « + 252)T 
*^*=2 70L Ja 



= _ 1M = -20.471 = -06M£. 
/. OaMS - 06M5 = -10.414 = -area of loop. 

J,t=3 
y dx gives in this case the difference of two areas. 
t=o 

To find the area OBC, where C has the same abscissa as M, viz., —4, 

we note that x = — 4 not only when £ = 2, which is the parameter of M 

whose ordinate is —8, but also when t = — 1, in which case y = 4, and 

this is the point C. The arc CO is traced from C to when £ varies from 

— 1 to 0. Hence 

OBC = f t= ° ydx = ^- [i 5 (30 1 2 - 175 1 + 252)]° = 6.53. 
•J t=— l 70 i 

163. Exercises (see the curves of Arts. 87 and 91). 

1. Find the area of the loop of the curve 

x = 10 (i 2 — 1), y = 10 (t*-t). 

2. Find the area of the loop of the curve 

x = t*-l, y = 5t*(t 2 - 1). 

3. Find the area of each loop of the curve 

x = io (* 3 -o, y = io (£ 4 -* 2 ). 

4. Find the area of each loop of the curve 

x = a sin d (1 + cos 0), y = a sin 2 6. 

5. Find the area of an arch of the cycloid 

x = a(e — sin d), y = a(l — cos 0). 

6. Find the area of each loop of the lemniscate of Bernoulli, 

x = a cos 4> Vcos2 </>, ?/ = a sin <£ Vcos 2 </>. 

7. Use the formula (3 r ) to find the area of each loop of the curve 

x = a sin cos 0, y = a sin 0. 

8. Find the area of the loop of the folium of Descartes (Art. 89), 

3 at 3 at 2 



y = 



In the foregoing problems it would be very difficult if not impossible 
to find the area from the x- and ^/-equations of the curve. In the problems 



234 



INTEGRAL CALCULUS 



164 



that follow, the area is to be found both from the x- and y-equation, and 
from the parametric equations. 

9. Find the area of each loop of the curve 

y 2 = x 2 (a 2 — z 2 ) 3 , or x = a sin d, y = a* sin cos 3 0. 

10. Find the area inclosed by the axes the line x = b, and the cate- 
nary 

y = ^(e a + e~ a ), orx = a log t, y = |^ + -j. 

11. Find the entire area between the z-axis and the witch of Agnesi, 



y = 



or x = a tan 0, y = a cos 2 0. 



x 2 + a 2 ' 

12. Find the area inclosed by the astroid 

xl -\- y\ = al, or x = a cos 3 d, y = a sin 3 0. 

13. Find the area of each loop of the curve 

a*y 2 + b 2 x & = a 2 6V, or x = a sin 6, y = b sin 2 cos 0. 

14. Given the curve 

(y -2 x 2 ) 2 = x 5 , or x = t 2 , y = V (t + 2). 

Find the area of the arch formed with the z-axis. 
Find also the area inclosed by the two branches of the curve and the 
line x = 4. 

15. Find the entire area within the closed curve 



ay = b 2 x 3 - x 4 



or 



¥ 



(1 - cos</>), y 



4a 



(1 — cos <t>) sin 0. 



164. The Derivative of Area: Polar Coordinates. Let U be 
the area BOP, bounded by the curve p = f(6), and two of its 

radii vectores OB and OP. Let p, 6 
be the coordinates of P. The ra- 
dius OB is any convenient starting 
place from which to measure U. As 
P traverses the curve, p turns about 
the pole, 0, and U plainly varies 
with 6. U is, in other words, a func- 
tion of 8 and has a derivative as 




to 6. We seek to determine this D U. 
Let 6 take an increment Ad =4- POP'. 



Draw the circular 



f 



§§165-166 AREAS 235 

arcs PN and P'M. Then Ap = NP' = PM, and AU = the curvi- 
linear triangle OPP' . It is now evident from the figure that 

area OPN < area OPP' < area OP'M, 
% P 2 Ad < AU < J(p + A P ) 2 A0, 

ip2< fd <i(p + Ap)*. 



or 
whence 




Passing to limits, we have 

(1) D 9 V = \p\ 

165. Areas: Polar Coordinates. From formula (1) of the 
preceding article we get by 

integration 

U = iJ P *dd + C. 

Then, by an argument similar 
to that of Art. 160, we get 

(2) u = if 6=y P *dd 

= if 6=y lf(e)] 2 do. 

Jd=p 

Example. 

Let us find the area of a loop of the lemniscate of Bernoulli (Art. 98, 
example 1) from its polar equation 

p 2 = a 2 cos 2 d. 
By formula (2), the area of the right-hand loop is 

U = hfJ P * do = | JVcob 2 tfe - ^(sin 2 8)\ = £ • 

166. Exercises (see the curves of Art. 99). 

1. Find the area of each loop of the curve p = a sin 3 6. 

2. Find the area inclosed by the curve p = 2 a cos 0. 

3. Find the area of each loop of the curve p = a sin 2 0. 

4. Find the area of the regions inclosed by the curve p = a sin§0. 

5. Find the area inclosed by the curve p = a sin 3 0. 

6. Find the area inclosed by the cardioid p = 2 a(l— cos 0). 



236 



INTEGRAL CALCULUS 



167 




7. Find the area of each loop of the curve p 2 = a 2 sin 6. 

8. Find the areas of the several regions inclosed by the curve 
p = a sin f 0. 

9. Find the areas of the several regions formed by each of the spirals 
of Art. 99, exercises 17, 18, 19. 

167. Lengths of Curves. In Chapter XIV, we found that 

D x s = Vl + y' 2 , and ds = Vl + y' 2 dx. 
Y From this we get by integration 

s= fVl+y'*dx + C, 

where s is the length of the arc 
measured from any convenient 
point, M, to the point whose ab- 
scissa is x. Therefore 
Arc AB = arc MB — arc MA = 

(Jvrfy~ 2 dx + c^ =6 - ( fvi+y^dx + c) _ , 

and if we now use s to denote the arc AB, we have 

(1) 8= f Vl +y' 2 dx, or s = f Vdx 2 + dy 2 . 

J a J a 

The second form of the formula is the more convenient when the 
curve is given by parametric equations. 

When the curve is given in polar coordinates, we have, by 
formulae V and V of Art. 102, 

D 9 s = Vp 2 + (D e p) 2 , and ds = V p 2 + p ' 2 dd, 
whence 

Vp 2 + P ' 2 dd. 
The process of finding the length of a curve is termed rectification. 



Example 1. Let us find the length of an arc of the upper branch of 
the semicubical parabola 

9 if = 4 x z , 

measured from the origin to the point whose abscissa is 3. From the 
given equation we have, for the upper branch, 

V = 5 x* , V' = *K Vl + y' 2 = Vl +x. 



§167 LENGTHS OF CURVES 237 

Hence, by formula (1) above, 

arc OP = fVl+z dx= \ |~(1 +a;)*Y= ^. 
Jo 3L Jo 3 

The parametric equations of this curve may be had by writing y = tx 
in the given equation. There results 



V , 9 
x = - t 2 , y 



t\ 



In order to obtain the length of the arc from 
these equations, we differentiate them and get 

27 



dx = — t dt, dy 



t 2 dt, 




Vdx 2 + dy 2 = ^Vi + 9t 2 tdt. 



Moreover, when x = 0, t = 0, and when x 
Hence by the second part of formula (1), 



3, i = 



V3 



arc OP = 7 C V ~W± + 9t 2 tdt=- n C v * V4 + 9 t 2 d(4: + 9* 2 ) 

4J 8J 



= ^[(4 + 9#] 



N3 



^(64-8) =| 



Example 2. Let us find the length of one turn of the logarithmic 
spiral p = ae bd from B = to d = x. We have 

P ' = a&e 6 *, V P 2 + P ' 2 = a VI + b 2 e™. 
Then, by formula (2), 

s - a VT+lf- Ce he dd = j Vr^¥(e b °y o 

J o 

= |vr+iT 2 (^ - 1). 

s is the heavy arc in the figure. 

When p = 0, = — go , and from this 
we know that between any point of the 
plane, as A, and the pole there are an 
infinite number of turns of the curve (the 
L part of the curve for which is negative 
The figure is drawn for b=VA^ayd^o is dotted in the figure). We can now 

show very easily that the total length of 




238 INTEGRAL CALCULUS §168 

this infinite number of turns between A and the pole is finite. We must 
integrate between = — oo (the pole), and = 0, (A). We have 

s = aVT+~^f° e^dd=^Vl+b 2 (e^)° ^VTTfc 2 . 

168. Exercises. 

1. Find the circumference of the circle from the equation 

y 2 + (x - a) 2 = a 2 , 
and also from the polar equation p = 2 a cos 0. 

2. Find the length of the loop of the curve 9 y 2 = x(x — 3) 2 . 

3. Find the perimeter of the astroid from the equation x%-\- ys = ah 
and also from the parametric equations x = a cos 3 6, y = a sin 3 0. - 

4. Let s in the catenary be measured from the point (o, a), and show 
that s 2 = y 2 — a 2 . Use, first, the equation 

and, second, the parametric equations 

x = alogt, y = |^ + -j. 

5. Find the length of one arch of the cycloid. 

6. Find the perimeter of the cardioid both from its parametric equa- 
tions 

x = a(2 cos 6 — cos 2 0), y = a(2 sin — sin 2 0), 

and from its polar equation P = 2c(l — cos 0). 

7. Find the length of the involute of the circle (Art. 96) 

x = a(cos + sin 0), y = a(sin — cos 0), 

from = to = 2 7T. 

8. Find the length of the spiral of Archimedes, p = ad, from = 
to = 2 x. 

9. Find the length of p = a sec from = to = j • 

10. Find the perimeter of a loop of the curve p = a sin 2 0. 

11. Find the length of the arc of the parabola between the vertex and 
the end of the latus rectum. Use the equation y 2 = 2 mx, and also 

m 
1 — cos 

12. Let us find the circumference of the ellipse 

x = a cos 0, y = b sin 0. 



168 LENGTHS OF CURVES 239 

Solution. We have 

ds = Vdx 2 + dy 2 = Va 2 sin 2 d + b 2 cos 2 d dd 

= Va 2 - (a 2 -6 2 )cos 2 0d0 = a Vl - e 2 cos 2 6 dd. 



Hence 



ds = 2a | VT 

z=— a *^ o 



e 2 cos 2 cte. 



The function a/1 — e 2 cos 2 cannot be integrated by elementary methods. 
That is, its integral is not one of the elementary functions: it is, in fact, 
an elliptic integral. An approximation to the value of this integral can 
be obtained by first expanding \A — e 2 cos 2 by the binomial theorem* 
and then integrating. We have 

, g2 g4 g6 K s>& 

(1 — e 2 cos 2 0)* = 1 — — cos 2 — — ■ cos 4 — — cos 6 — — — cos 8 6. . . . 
.2 8 lo 128 

Integrating term by term by the aid of a table of integrals, and then 
substituting the limits, we have 

circumference ) _ p [~i _ f? _ 3^ _ 5e 6 _ 
of ellipse j-^ xa L 4 64 2 56 ' * J" 

13. Find the circumference of the ellipse 

9x 2 + 25y 2 = 225. 

14. Find the circumference of the ellipse 

3x 2 + 4y 2 = 12. 

* In Chapter XXX it will be shown that this expansion is legitimate. 



CHAPTER XXIII 

THE DEFINITE INTEGRAL AS THE LIMIT OF A SUM OF 
INFINITESIMAL PRODUCTS 

169. The Fundamental Theorem of the Integral Calculus. 

Let f{x) be real, single-valued, and continuous throughout the 
interval a rE x ~~ b. Divide this interval into n subintervals by 



A a &x, ax, Aa?„ wl 



%2 



the points x h x 2 , . . . x n - h and let the lengths of these subinter- 
vals, which may or may not be equal, be denoted by Aa, Ax h 
Ax 2 , . . . A£ n -i. Form the sum of products 

f(a)Aa+f(x 1 )Ax 1 +f(x 2 )Ax 2 + . . . /(:c B _i)Aa; n _i, 

b 
and denote it by V/(x)Az. Now let each subinterval be sub- 

a 

divided, and a new sum of products formed, and let this process 
go on indefinitely. The single restriction upon the method of 
forming the subintervals is that each of them must have the limit 0. 

b 

The number of products in the sum £,f(x) dx increases indefinitely, 

a 

while each product, f(Xi)Axi, decreases indefinitely, that is, becomes 
infinitesimal (Art. 10). Now the theorem that we are about to 

240 



169 



THE DEFINITE INTEGRAL 



241 



prove is that the sum ^f(x) dx, has a determinate limit. Expressed 

a 

in symbols, the theorem is, 

b 
as n = oo , V/(z)Az = a definite limit. 

a 

This is a purely analytic theorem, and admits of a purely analytic 
proof, but we shall give here a geometric proof of it. 

Proof. We represent the values of f(x) by ordinates of a curve, 
y = }(x), and shall suppose first that /(a;) is positive and increasing 
throughout the interval a to h. With the several ordinates as 
altitudes, we construct a series of rectangles as shown in the figure. 
Because f(x) is, by hypothesis, an increasing function, each of 
these rectangles lies between 
the 'curve and the x-axis. 
Each product /(£»)Az» is the 
area of a rectangle of base 
Ax» and altitude /(#»), and 

the sum^X/(x)A^ is the sum 

a 

of the areas of all these 

rectangles. Our theorem is 

proved, then, as soon as it is proved that this sum of rectangles 

has a limit. Now the figure leads us to suspect at once, and 

indeed to be reasonably certain, that this sum of rectangles has as 

a limit the curvilinear quadrilateral aABb; that is, that 

b 

lim X /0&)Aa5 = area of aABb. 

Of this, which is so nearly self-evident, we shall now give a rigorous 
proof. 

In addition to the interior rectangles of the figure above, we 
construct, as in the accompanying figure, a series of rectangles 
extending partly above the curve, such as Nxt, and which we term 
exterior rectangles: Each Ax is the base of one interior and one 




a x-l a? 2 



242 



INTEGRAL CALCULUS 



169 



exterior rectangle. Let E be the sum of the areas of the exte- 
rior rectangles. It is geomet- 
rically obvious that 



(a) E > aABb > ^f(x)Ax. 

a 
b 

Moreover, E — 2 fix) Ax is 

a 

the sum of the small rec- 
tangles, such as NR, each of 
which is the difference be- 
tween an exterior and an in- 
terior rectangle on the same base. These small rectangles have 
different altitudes, but the sum of their bases is always b — a. 
Let h n be the greatest of their altitudes for a given value of n. 
Then 




(b) 



E-j?f(x)Ax<h n (b-a), 



and both members of this inequality are positive, and the inequal- 
ity holds for all values of n. Moreover, as n == bo, h n = 0, and 
h n (b — a) = 0. Consequently 

lim \E - j? fix) Ax) = 0, whence \imE= lim ^. fix) Ax. 

b 

Since E andV/(x)Arc have the same limit, it follows from (a) that 

a 

this limit is aABb. Therefore we have proved 

b 
(I) \im y £ l f(x)Ax = aABb. 

«-°° a 

Problem. Prove (I) when f(x) is a decreasing function. Prove it for 
an interval in which fix) is sometimes increasing and sometimes decreas- 
ing. Consider also the case in which fix) is negative throughout the 
interval. 

We have already learned, Art. 160, that 

area of aABb = J fix) dx. 



§169 

Therefore 

(II) 



THE DEFINITE INTEGRAL 



lim V/(x)Ax = f f(x)dx. 



243 



Not only, then, has the sum of infinitesimal products]^ f(x) Ax 

a 

a finite and determinate limit, but this limit is a definite integral. 

These results are of such importance in the Integral Calculus that 

we restate them in the 

Fundamental Theorem. Let f(x) be real, single-valued, and 

continuous throughout an interval a = x = b, and let this interval be 

subdivided into n smaller intervals by the points Xi, x 2 , . . . x n -i, 

b 

and let the sum of products be formed, ^f(x)Ax. Now let n, the 

a 

number of subintervals, be increased indefinitely by repeated subdi- 
vision of each subinterval, the process of division being carried on in 

such a way that each subinterval has the limit 0. Then, although 

b 

each product f(x)dx has the limit 0, the sum ^f(x)dx has a limit 

a 

which is the definite integral, 



f(x) dx. That is, 



x 

lim *%f{x)dx= I f(x)dx. 

n =oo a Ua 

It is evident from our defini- 
tion and from the accompany- 
ing figures that 



Jf(x) dx = — I f(x) dx, 
a Jb 



and 




f b f(x)dx+ Pf(x)dx+ f d f(x)dx= f d f(x)dx, 

*J a Ub %J c *J a 



and that these hold true whatever be the order of the limits, a, b, 
c, d. 

Problem. Show how the proof of the fundamental theorem depends 
upon the continuity of fix). 



244 INTEGRAL CALCULUS §170 

We have already stated that, although the foregoing proof of 
the fundamental theorem is geometric, the theorem itself is a 
purely analytic one. This means that the theorem holds true 
when the values of f(x) are something else than the ordinates of 
a curve, and when therefore the limit of the sum of products is 
something else than an area. We shall have many examples of 
this. 

It may be remarked here that this property of being the limit 
of a sum of products of the form f(x)dx could be taken as the 
definition of the definite integral. Indeed, to the originators of 
the Calculus the definite integral first presented itself as the limit 
of such a sum of products, and from this circumstance arose the 

name integral and the symbol / ; for, the word integral means 

the whole regarded as the sum of its parts, and the symbol / is 

merely an elongated S which originally stood for the word sum. 
In viewing the area under a curve as the limit of a sum of rectan- 
gles, each of the component rectangles is termed an element of 
area. 

170. Area of a Sector. In Art. 165 was derived the formula 
for the area in polar coordinates of the sector OBC. It was found 

that 

area OBC = \ 

We shall here show how this 
formula, may be gotten by 
means of our fundamental 
theorem. 

We first divide the sector 
OBC into n smaller sectors, 
such as OPP', and then con- 
struct, from as center,"a cor- 
responding set of n interior 
circular sectors, such as OPN, each having one vertex on the curve. 




§170 THE DEFINITE INTEGRAL 245 

Next we subdivide the small sectors OPP', construct a new set 
of interior circular sectors, and continue this process indefinitely, 
taking care that the successive subdivisions shall be so carried 
on that each small sector shall have the limit 0. We shall now 
show that 

area OBC = lim 2) OPN. 

n=co p 

To this end we produce the radii vectores beyond the curve, and 
construct from as center a set of n exterior circular sectors such 
as OP'M. It is geometrically evident that for each value of n, 

(a) 5) OPN < OBC < 5* OP'M. 

Let $ denote the sum of small curvilinear quadrilaterals such 
as PNP'M. Then 

j?OPN -j?OP'M=S. 

Let R be the greatest radius vector in OBC. With R as radius, 
and center at 0, we strike the circular arc EE' = R(y — /3). The 
sum of the exterior bases, P'M, of the curvilinear quadrilaterals, 
PNP'M, is plainly less than R (7 — /3), and if h n be, for a given 
n, the greatest of their altitudes, we have 

< S < R(y - 0)h». 
R and 7 — (3 are constants, while h n = as n = 00 . Therefore 

lim£ = 0, and lim 2) OPN = lim J) OP'M. 

Consequently, by virtue of (a), 

area OBC = lim 2) OPN. 

It is now a very simple matter to show that area OBC can be 
expressed as a definite integral. We have 

PN= P Ad, whence OPN=i P 2 k9. 
Therefore 

area OBC = |limXp 2 A0. 



246 



INTEGRAL CALCULUS 



§§ 171-172 



Now in the last sum p 2 is a function of 6, and therefore the sum of 
products is formed in the way contemplated by our fundamental 
theorem. Hence, by that theorem, 

area OBC = JlimV P 2 A0 = \ f \ 2 dd. Q. E. D. 

In this case the element of area is § p 2 dd. 

171. Proof that Every Function has an Integral. By aid of 

our fundamental theorem we can now prove the statement made 
in Art. 122, that every function has an (indefinite) integral. 

Let f(x) be real, single-valued, and continuous throughout the 
interval a = x~x x . We shall prove that there exists a function 
whose derivative is j{x\). Form the function F(xi) defined by the 
equation 

F(x 1 )=limj>f(x)dx. 

From our fundamental theorem we know that such a function 
can always be formed. 

We know that F(xi) is repre- 
sented geometrically by the area 
aABxi, and that Bxi = f(xi). 
Moreover, in Art. 159, it was 
proved that D Xl F(xi) = f(xi), where 
Xi is variable, or, dropping sub- 
scripts, 
D x F(x)=f{x). 

Therefore, since F(x) can always be constructed, there always 
exists a function whose derivative is f(x) . In other words, every 
(continuous) function has an integral. 

172. The Problem of Motion. In Art. 147 we obtained the 
formula for the distance traversed by a moving body, viz., 

s = I vdt + C, 
where v is a function of t. The presence of the unknown constant 



F(xJ 



f<&) 



§172 THE DEFINITE INTEGRAL 247 

C means that the initial conditions of the problem are as yet 
undetermined; that is, that the point from which s is^measured, 
and the instant from which t is counted, are as yet undetermined. 
If s b and s a denote the values of s when t = b and t = a respectively 
(b > a), then s b — s a is the distance traversed in the time interval 
from t = a to t = b, and 



Sb — S a 



{f vdt+c L-{f vdt+c h 

Writing for sake of simplicity s for s b — s a , we have 

s = / vdtj 

a formula which expresses s, the distance traversed from t — a to 
t = b, as a definite integral. In obtaining this formula we have 
followed the method used in Chapter XXII, in the case of areas 
and lengths of arcs. It is interesting to see how it may be obtained 
by means of our fundamental theorem of Art. 169. 

Divide the time interval a to b into subintervals by the points 
of time ti, Uj h, . . . t n -\, and let Aa, At h A£ 2 , . • • A£ n _i, be the 
lengths of these subintervals of time. Let v a , vi, v 2 , . . . v n -i, 
be the (instantaneous) velocities at the beginning of these time 
intervals. Then v a ka is the distance the body would have moved 
in the time interval A a had its velocity been uniform and equal to 
v a throughout this interval; so, too, vi&ti is the distance the body 
would have moved in the time interval A£i had its motion contin- 
ued uniform with the velocity v\ throughout this interval; and so on. 

b 
Hence X^Ai is the distance the body would have moved in the 

o 

time interval a to b had its motion been made up of a succession 
of uniform motions through successive small intervals of time. 
And it is obvious that the actual motion may be regarded as made 
up of an infinite succession of uniform motions through successive 
infinitesimal time intervals, or, more accurately speaking, that the 
actual motion is the limit of such a succession of uniform motions. 



248 



INTEGRAL CALCULUS 



§173 



In other words, s, the actual distance traversed in the interval 
from t = a to t = b, is given by the formula 



= lim]Tz;AZ. 



Now since v is a function of t, the second member of this equation 
is a sum of products to which our fundamental theorem applies, 
and we have 



n 



lim Z/VAt = I 

n = x ~~ 7 U a 



vdt. 



Q. E. D. 




It is interesting to note that if we represent time by abscissas 
W and velocities by ordinates, so that 

v = f(t) is "the equation of the curve 
in the figure, which may be termed 
the velocity curve, then s is the 
area aABb. 

Also, D t v = acceleration = slope 
a t\ *2 K tn-ib of the velocity curve. 

Query. What do the small rectangles in the figure represent ? 

173. A Theorem of Infinitesimals. The application of the 
fundamental theorem is greatly facilitated by the use of Duhamel's 
Theorem, a theorem of infinitesimals which corresponds to that 
proved in Art. 103. It is as follows: 

Duhamel's Theorem. In finding the limit of the sum of an in- 
finite number of positive infinitesimals, if this limit is finite, each 
or any infinitesimal of the sum may be replaced by any other infini- 
tesimal whose ratio to the one replaced has the limit unity. 

Expressed in symbols, this theorem is: 
If on, a 2 , . . . oL n are positive infinitesimals, and if 



then 



lim— = 1, lim— = 1, . . . lim— = 1, 

Oil «2 Oi n 



lim (|8i + /3 2 + • • • /?») = Um (ai + a 2 + . . • a n ), 

n=oo n=°o 

provided this limit is finite. 



§173 



THE DEFINITE INTEGRAL 



249 



Proof. It is an easily proved theorem of algebra that if there 
be given a series of n positive fractions, 

01 02 0n 

R R 

and if — be the smallest of these and — the largest, then 



(a) 



0r 

a r 



+ 02 + 



+ 



a a 



Oil + «2 + • • • + OLn 

Let the student give the proof of this. 

This theorem applies in the present case, and, therefore, taking 
the limits in (a), we have 

lim ft slim /5i±ft+ - 

n = oo Ot T 

Hence 



n iooai+a!2+ • • • &n n = ooOC a 

But, by hypothesis, the first and last limits are each 1 



lim-f 

n=oo 2/« 



1, 



and from this it follows that 

lim (0i + 02 + . . . + 0n) = lim ( ai + a 2 + 



. +a»). 



As an example of the application of this theorem, let us find the 
area in polar coordinates of the sector OBC. 

We divide OBC into smaller sectors, subdivide these into still 
smaller ones, and so on ad infinitum, taking care that the method 
of subdividing be such that the small 
sectors shall each have the limit 0. 
Let n be the number of small sectors 
after any subdivision and OPP' the 
type. Since OBC is the sum of all 
these n sectors, it is also the limit of 
their sum as n is indefinitely increased; 
that is 

(b) area OBC = lim % OPP'. 

n = oo 

In the second member of this equation we have the sum of an 




250 INTEGRAL CALCULUS §173 

infinite number of positive infinitesimals, and the limit of this 
sum is finite. Hence, by our theorem, each of the infinitesimal 
sectors OPP' may be replaced by another infinitesimal, provided 
their ratio has the limit 1. On each component sector OPP' we 
now construct, as in Art. 170, an interior circular sector OPN. 
We shall show that OPN is the infinitesimal which may replace 
OPP' in (b). Construct on OPP' the exterior circular sector 
OP'M. It is obvious that 

OPP' OP'M 
OPN < OPP' < OP'M and 1 < ^^ < —^ , 

whence 

, v , < r OPP' < .. OP'M 

(c) 1 = ^opn = ^Zopn' 

Now 

OPN = ip 2 Ad = i P 2 dd and OP'M = \ (p + Ap) 2 dd. 
Hence 

OPM (p + Ap) 2 
OPA^ p 2 

and 



lim 



OPN 



-ft, (,+*£) -1. 

n=oo \ p / 



OPP' 

Consequently, by reason of (c), lim npAr = 1, and, by Duhamel's 

Theorem each OPP' in (b) may be replaced by the corresponding 

OPAf, that is, by \ p 2 dd, so that we have 

y y 

area OBC = lim Y J p 2 dd = \ lim V p 2 d0. 

n=oo o n=oo o' 

Finally, p 2 is a function of 6, and we may therefore invoke our 
fundamental theorem of Art. 169, and thus get the formula 

as in Arts. 165 and 170. 

Let us apply our fundamental theorem to obtain the formula 
for the length of an arc of a curve 

s = f Vl +y'*dx. 



s 



area OPC = \ I P 2 dd, 



§173 



THE DEFINITE INTEGRAL 



251 



Let s be the length of the arc AB, and let this arc be divided 
into n small arcs whose lengths are denoted by As , Asi, As 2 . . . . 
Then, whatever value n may Y 
have, 

6 

s = 5) As, 

a 

and when n is increased indefi- 
nitely by repeated subdivision of 
the small arcs, we have 




(d) 



= lim ]T As. 



We have the well-known relations 

As _ ds_ ^_^ s . A 
Al = Ax dx' Ax ~ dx 

where e is infinitesimal. Hence 

As 1 . dx . .. As 

- J - = l + €- r -, and lim -T- = 
ds ds ds 



+ edx, 



1. 



Therefore, by our theorem of infinitesimals, each As in (d) may be 
replaced by the corresponding ds, and we have 

b b 

s = lim V ds = lim TV1+ y' 2 dx. 

„==o n^oc^ 

But Vl + 2/' 2 is a function of x, and therefore to the last sum- 
mation our fundamental theorem applies, so that 

s= f VI + y' 2 dx, 

and this is the formula sought. 
Remark. It has been proved here incidentally that 

.. increment of the function _ 
differential of the function 



Query. What does J) ds represent geometrically? 



252 



INTEGRAL CALCULUS 



§173 



Problem 1. Let ABCD be an area bounded by a circle, its involute, 

and two positions AB and CD of the 
generating line. Show that \a 2 B 2 dd 
may be taken for the element of area, 
and that 

ABCD = ^(d 2 *-eS). 
6 

Problem 2. There is a wall inclosing 
a circular space of radius a. To a ring 
on the outside of the wall a horse is 
attached by a rope. Over what area 
can the horse graze when the length 
ira. 




of the rope is 



. to, f — — - I 
2 2 



2 ira? 



Over what area can he graze when 
the length of the rope is b where 
b <7ra? 

The student should now work some of the exercises of Chapter 
XXII in the light of the principles of the present chapter. 



CHAPTER XXIV 



SURFACES AND SOLIDS OF REVOLUTION 



174. Definitions. When a plane is revolved about one of its 
fixed right lines as an axis, every point in the plane describes a 
circle, every right line or curve describes a surface of revolution, and 
every portion of the plane bounded by lines and curves describes 
a solid of revolution. The surface of revolution described by a 
right line is a cone or a cylinder. 

When a conic revolves about a principal axis, the surface or 
solid of revolution generated is termed an ellipsoid of revolution, 
a hyperboloid of revolution, or a paraboloid of revolution, according 
to the species of the conic. The sphere is a particular case of the 
ellipsoid of revolution. When the ellipse is revolved about its 
major axis, the resulting ellipsoid is sometimes termed a prolate 
spheroid, when about its minor axis, an oblate spheroid. When 
the hyperbola is revolved about its transverse axis, the resulting 
hyperboloid is termed an hyperboloid of two sheets, when about 
its conjugate axis, an hyperboloid of one sheet. 

175. Volumes of Solids of Revolution. Let A B be an arc 

of the curve y = f(x), and let V 
be the volume of the solid gen- 
erated by revolving about OX the 
curvilinear quadrilateral aABb. 
We seek an expression for V. 

To fix ideas, we suppose that 
f(x) is an increasing function 
throughout the interval a to b. 
It is assumed, of course, that f(x) 
is single-valued, continuous, and 
real throughout this interval. Divide the interval a to 6 into n 

253 




254 INTEGRAL CALCULUS §175 

subintervals, erect ordinates at the points of division, and complete 
the interior rectangles as shown in the figure. As the plane re- 
volves about OX, each of the curvilinear quadrilaterals, such as 
PQST, generates a solid of revolution whose volume we denote 
by Vq, and it is obvious that V is the sum of the volumes Vq what- 
ever be the value of n. Now let n increase indefinitely by re- 
peated subdivision of each subinterval of the a>axis, and we have 
of course 

b 

(a) V = lim V Vq. 

We shall now show that in the second member of this equation 
each Vq may be replaced by Cp, the volume of the cylinder gen- 
erated by the corresponding interior rectangle PRST. On the 
base ST or Ax, construct the exterior rectangle P'QST, and denote 
the volume of the cylinder which it generates by Cp>. Then it 
is geometrically evident that 

C P < V Q < C p > and 1< -^ < ~- 

Lp Lp 
Now Cp = iry 2 Ax, and Cp> = ir(y + Ay) 2 Ax, and therefore 

(■+!?)'• 



C P y 2 



and 1 ^lim J? ^ lim ( 1 + ^Y = 

n =oo Lp Ay 



Hence lim -^r = 1, and, by DuhameFs theorem of infinitesimals, 

n =oo Cp 

each Vq in (a) may be replaced by the corresponding Cp or by 
Ty 2 Ax. Consequently we have 

b b 

V= lim 2} iry 2 Ax = 7r lim V 2/ 2 Az. 

n=oo a 7l = oo ■ 

Our fundamental theorem is applicable here, and we have 

(A) V=7r£y 2 dx, 

which is the formula for the volume of a solid of revolution. In 



§176 



SURFACES AND SOLIDS OF REVOLUTION 



255 



a similar manner it can be shown that 
the volume of the solid generated by 
revolving about OY the curvilinear 
quadrilateral a'AEb' is 



(A') 



■r 



x 2 dy. 




The elements of volume in (A) and (A') 
are respectively 

iry 2 Ax and irx 2 Ay, or iry 2 dx and irx 2 dy. 
176. Areas of Surfaces of Revolution. Let S be the area of 
the surface generated by revolving the arc AB about OX. We 
seek an expression for S. 

Divide the arc AB into n small arcs and draw their chords. 
As the plane revolves about OX each of these chords, such as c, 
generates the convex surface of the frustum of a right circular 

cone, and if S c denote the sur- 
face generated by c, the sum of 
the surfaces generated by all 
the chords may be denoted by 

b 
a 

Now subdivide each of the 
small arcs, draw their chords, and take the sum of the convex 
surfaces that these chords generate, and continue this process in- 
definitely, taking care that each small arc shall have the limit 0. 
As n is thus indefinitely increased, the sum of the surfaces gener- 
ated by the chords c will approach as a limit the surface generated 
by the arc AB; that is, 



Y 

\ 


\ 


\y 


y 


y+Ay 


B 









Ax 




X 




c 


i a 


n 


I 


? 



limj?S c . 



The area of the convex surface of the frustum of a cone is equal to 
the product of the slant -height by the circumference of the circu- 
lar section made by a plane midway between the bases. In the 
figure the slant height of S c is c and the circumference of the cir- 



256 INTEGRAL CALCULUS §176 

cular section midway between the bases is 2 tt (y + § Ay) . Hence 
S c = 2 7r (y + | Ay) c. Consequently 

(a) S = \im^.27r(y + i Ay)c=2Tr\imy (y + % Ay)c. 

Since c is infinitesimal, the product (y + } Ay)c is an infinitesimal 
product. Consider .the ratio of this infinitesimal product to the 
infinitesimal product yds, 

(y + jAy)c _/^ , jAy\ c As 

and 

(y + iAy)c _,._/, , lAy\_ i: _ c _ i; _As 



yds V y / As ds 

iim ^±iM« = lim (i + IM . lim c . lim 

yds \ y I As 



Each of the three limits in the second member of this equation 
is 1 (see Remark in Art. 173) and consequently 

lim (y + f y) ° = 1. 
yds 

Therefore, by Duhamel's Theorem, Art. 173, each infinitesimal 
product in (a) may be replaced by the corresponding yds, so 
that we have 

6 

S = 2 t lim V yds. 

Applying the fundamental theorem, we get the formula for the 
area of a surface of revolution, 



(B) 



= 271-/ yds, 



or, because ds = VI + y' 2 dx 1 

(B) S=2<ivj b y^l+y' 2 dx. 

In a similar manner, it may be shown that the formula for the 
area of the surface generated by revolving the arc A B about OY is 

(B r ) S = 2tt P xds = 2tt f b xVi+ y ' 2 dx. 

U a' *J a' 



X 



§176 SURFACES AND SOLIDS OF REVOLUTION 257 

In (B) and (B') the elements of surface are respectively 

2 wy ds and 2 irx ds. 

Example 1. Find the volume and surface generated by revolving 

about OX the ellipse^ + f; = 1. 
a 2 b 2 

Solution. We need consider only that part of the curve which lies 

above OX. 

1st. The Volume. From the equation of the curve we have 

y 2 = ~ i (a 2 -x 2 ). 
a 2 

Then 

V = 7T y 2 dx = — ) (a 2 - x 2 )dx = — la 2 x - - 

J - a a~ J -a a 2 V 3/- a 



^ & 7 4 s\ 



|.a& 2 . 



When b = a, F = | A-a 3 , the well-known formula for the volume of a sphere 
of radius a. 
2d. The Surface. We have 

, -bx , 1 . , 2 a 4 -c 2 z 2 a 2 -e 2 z 2 

^ = — / , and 1 + ?/' 2 = -— = — • 

a Va 2 - x 2 « (a - z 2 ) a 2 - & 

Also, 2/ Vl + y' 2 = ? V^ 2 ^^ ^ f ~ eV = - (a 2 - e 2 * 2 )^ 

a Va 2 - Z 2 a 

Therefore 

2rb r + 

-a ae - 



fi = 2tt f yVl+y'*dx = — f (a 2 - e 2 x 2 )*dex 

= — [ex Va 2 - e 2 x 2 + a 2 sin- 1 — l + ° 
aeL a J- a 



= — [a 2 e Vl - e 2 + a 2 sin" 1 e] 
and, finally, 

sm — ^ e 
When 6 = a, e =0, and = 1, and we have the well-known formula 

for the surface of the sphere of radius a, 

£ = 47ra 2 . 

Example 2. Find the volume and surface of the ring generated by 
revolving the ellipse about a line in its plane parallel to the major axis 
and at a distance k from it. 



258 



INTEGRAL CALCULUS 



§176 



We choose for OX the axis of revolution and for OY the minor axis of 
the curve. The equation of the curve is then 
Y 




b 2 x 2 + a 2 (y - k) 2 = a?b* 



whence 



k ±- Vo 2 " 
a 



We suppose throughout that k = b. 

1st. The Volume. The required vol- 
ume, V, is the volume generated by 
the revolution of the plane figure 
C'A'BAC minus the volume generated 
by the revolution of C'A'B'AC. De- 
X noting these two volumes by Vb and 
Vb', we have V = Vb — Vb'. 
Now 



y B==7r f +a (k + b Va2 _ 
J -a \ a 

= tt J*" f> + ^ (a 2 - x 2 ) + ™ VoJ^lf] dx, 
Vb> = tt f +a (k-- a VgT^A* 



,+a 



dx 

b 2 , , 0N 2 kb 



= *( [k 2 + - (a 2 -x 2 )- — VoJ^ 2 ] dx. 
J-a La 2 a J 

Hence 

v = iM r+« y/fzr&te = 2Mr v^3^i + a2 sin - 

a ./_o a L 



aJ_ 



2 irkb ( irQ 2 . ttQ 2 

a V 2 + 2 



)- 



2 7r 2 /cafe. 



This may be written 



V= 2irk*irab. 



Now ?ra6 is the area of the ellipse, and 2 wk is the length of the circumfer- 
ence described by its center. Therefore the ring is equal in volume to a 
right cylinder whose base is the ellipse, and whose altitude is the distance 
revolved through by its center. 

The volume may also be obtained from the parametric equations of 
the curve. Using the same axes of coordinates as before, we have 



x = a cos 0, y = k + b sin 0. 



§176 SURFACES AND SOLIDS OF REVOLUTION 259 

Now we may restrict to vary from to w only, and use the equations 
just written as the equations of the arc ABA' and 

x = a cos d, y = k — b sin 

as the equations of the arc AB'A', each arc being traced in the negative 
direction of the z-axis. Then, since dx = —a sin dd, we have 



7 B = w a C (k + 6 sin 0) 2 sin d0, Vb' = -na J (/c — 6 sin 0) 2 sin 



dd. 



and 



7 = 7 5 — y B - = 4 7r&a6 J ' sin 2 Odd = 2 wkab 1 — sin cos ) , 



whence 7=2 7r 2 fca& 

as before. 

Or we may let vary from to 2jr, and then the equations 

x = a cos 0, 1/ = fc + 6 sin 

will represent the whole circumference, the point {x, y) tracing the curve 
in the direction A, B, A', B\ A. Then 



V = iva f *" (k + b sin 0) 2 sin dd = wa C * (7b 2 + 2 kb sin + 6 2 sin 2 0) sin 



.2x /»2x 

(/c + b sin 0) 2 sin ddd = wa J 

t6 2 ~\ 2ir 

— A; 2 cos + kb(d — sin cos 0) + — cos 3 — b 2 cos =2 x 2 A;a6. 



In this example the last method involves a more difficult integration than 
does the one preceding it. 

2d. The Surface. The required surface, S, is the sum of the surfaces 
Sb and Sb' generated by the arcs ABA' and AB'A'. 



Sb = 2 wf^ (k+- Va 2 - xAds and Sb> = 2 tt f + " 6b - - Va 2 - aA 
Therefore £ = 2 dfc • 2 J ° ds. 



In Art. 168, exercise 12, it was found that this last integral in the case 
of the ellipse cannot be calculated by elementary methods. We know, 
however, that this integral gives the circumference of the ellipse, and 
consequently 

S = 2 wk (circumference of ellipse). 

Therefore the surface of the ring is equal to the convex surface of the cylinder 
mentioned on the preceding page. 



260 INTEGRAL CALCULUS §177 

177. Exercises. 

1. Find the volume and surface generated by revolving a circle about 
a diameter. 

2. Find the volume and the surface cut from a sphere by two parallel 
planes. 

3. From the parametric equations of the ellipse find the volume and 
surface generated by revolving the curve about its minor axis. 

4. Find the volume generated by the arc of the cubical parabola 
y = ax z included between the origin and the point whose abscissa is b, 
when revolved, (1) about OX, (2) about OY. 

5. Find the volume generated by revolving about OX the oval of the 
curve 9 y 2 = x(x 2 — 3). (Art. 87, exercise 5.) 

6. Find the volume generated by revolving about OX the curve 
a 2y2 = 52^3 _ x i (^ r ^ 87^ exercise 7.) 

7. Find the volume and surface generated by revolving about OX 
the astroid xi + y% = a». 

8. Solve 7, using the parametric equations of the astroid. 

9. Find the volume and surface of the ring generated by revolving 
the astroid about a normal at a cusp. What cylinder has the same 
volume and the same convex surface? 

10. Find the volume and surface of the ring formed by revolving a 
circle about any line of its plane which does not cut the circumference; 
and also find the volume and surface formed by revolving the circle about 
a tangent. To what cylinders are these rings equal in volume and convex 
surface? 

11. A circle is revolved about one of its tangents. Find the volume 
and surface generated by each of the semicircles formed by a diameter 

parallel to the tangent. 

12. A mechanic places in a lathe a cylinder 
of wood of length 4 a and of base radius 6. 
He then turns the block of wood so that the 
contour of the convex surface shall have the 
form of an ogee, consisting of two semicircles 
as shown in the figure. Find the volume and 
convex surface of the solid thus formed. 

13. A loop of the lemniscate 



x = a cos <p Vcos 2 <f>, y = a sin <\> Vcos 2 <f> 

is revolved, (1) about OX, (2) about OY. Find the areas of the surfaces 
thus formed. 

14. A closed curve which has an axis of symmetry is revolved about 



§177 SURFACES AND SOLIDS OF REVOLUTION 261 

an axis in its plane parallel to the axis of symmetry but not cutting the 
curve in real points. Show that the volume and surface of the ring 
generated are equal to the volume and convex surface of a right cylinder 
whose base is the area inclosed by the given curve, and whose altitude 
is the length of the circumference described by any point in the axis of 
symmetry. 

15. Apply the principle of exercise 14 to find the volume and surface 
of the ring generated by revolving the astroid about the line joining two 
successive cusps. 

16. Apply the principle of exercise 14 to find the volume of the ring 
formed by revolving the lemniscate: (1) about a double tangent; (2) about 
a tangent at a vertex. (Art. 163, exercise 6.) 

17. Apply the principle of exercise 14 to find the volume of the ring 
formed by revolving the curve p = a sin 2 0: (1) about one of its double 
tangents; (2) about the tangent at the vertex of one of its ovals. (Art. 
166, exercise 3.) 

18. Find the volume and surface generated by revolving about OX 
an arch of the cycloid. 

19. Find the volume and surface generated by revolving about OY 
that arch of the cycloid which passes through the origin. 

20. Find the volume and surface generated by revolving an arch of 
the cycloid about the tangent at the highest point. Use the equations 
of Art. 93, problem 2. . 

21. Find the volume and surface generated by revolving an arch of 
the cycloid about a normal at the highest point. 

22. Find the volume generated by revolving about OX an oval of the 
curve y 2 = a 2 sin x. 

23. Find the volume and surface generated by revolving about OX an 
arch of the sine curve y = sin x. 

24. Let the sine curve be revolved about one of its multiple tangents. 
Find the volume generated by the arc included between two successive 
points of tangency. 

25. Use the principle of exercise 14 to find the volume of the ring formed 
by revolving OX and the nth. arch of the sine curve about OY. Find 
the volume of the solid formed by revolving OX and the first n arches 
about OY. 

26. Find the volume generated by each oval of the curve 

4 2/2= ( X 2_ 1)2(4_ X 2) 

when revolved, (1) about OX, (2) about its multiple tangent. (Art. 87, 
exercise 21.) 



262 INTEGRAL CALCULUS §178 

27. . Find the volume generated by a single loop of the curve 

ay + b 2 x* = a?b 2 x 4 , 
or x = a sin d, y = b sin 2 cos 

when revolved, (1) about OX, (2) about OY. (Art. 87, exercise 25; 
Art. 91, exercise 12.) 

28. Find the volume and surface generated by revolving about OX an 
arc of the catenary, 

V = ^((P + e~"J, 
or x = a log J, y = Tift + z Y 

29. Find the total volume generated by revolving, (1) about OX, and 
(2) about OY, the witch of Agnesi, 



or x = a tan 6, y = a cos 2 6. 



X* -+- (T 

(Art. 87, exercise 15; Art. 91, exercise 19.) 

30. Find volume generated by each loop of the curve 

x = a cos 6 (1 + cos d), y = a cos sin 0, 

when revolved about OX. (Art. 91, exercise 24.) 

31. Find the volume generated by each loop of the curve 

x = a sin cos d, y = a sin 0, 

when revolved about OX; about OY. 

32. Find the volume formed by revolving the curve 

x = a sin (1 + cos 0), y = a sin 2 0, 

(1) about OX, (2) about OY. (Art. 91, exercise 25.) 

33. Find the volume generated by revolving the curve of exercise 32 
about its double tangent. 

34. Find the volume generated by revolving, (1) about OX, and (2) 
about OY, the loop of the curve 

x = t 2 - 1, y = 5t*(t 2 - 1). 
(Art. 91, exercise 3.) 

178. Solids and Surfaces of Revolution. Polar Coordinates. 

The necessary f ormulse are obtained by transforming formulae (A) , 
(A ; ), (B), (B') of Arts. 175 and 176 to polar coordinates by means 
of the equations 

x = p cos d, y = p sin 8, ds = V p 2 + p ' 2 dd. 



§178 SURFACES AND SOLIDS OF REVOLUTION 263 

By this means we get 

y 2 dx = w I p 2 sin 2 6 d(p cos 0), 

S = 2tt f b yds = 2tt f p sin B V p * + p '* dO, 

t/o t/a 

V' = tt Tx 2 % = tt f p 2 cos 2 0d(psin0), 
5' = 2tt Txds = 2tt f p cos Vp 2 + p' 2 d0. 

t/a' t/a 

Example. To find the volume and surface generated by revolving 
about OX the curve p = a cos 2 0. y 

By the foregoing formula, 

V = — 3 wa? L cos 6 sin 3 dd 

2 

= 3 7ra 3 f (cos 8 — cos 6 d)d cos 

IT 

= 3 xa 3 [i cos 9 — £ cos 7 0J = — - • 

o -^1 

Therefore the volume generated by both loops is 2 V = 
Further, 




4xa 3 
21 



S = 2 Tra 2 f a/4 - 3 cos 2 cos 3 d cos 0. 



This integral is of the form f\/4: — 3u 2 u 3 du and can be integrated 
by the table of integrals. (See also exercise 14, Art. 135.) We find that 

J\/4- 3 v? u*du= - -^ (8 + 9w 2 ) (4 - Su 2 )\ 
135 

Therefore, since u = cos 0, 

S = ^g [(8 + 9 cos* 0) (4 - 3 cos' ,)*£- |ff (64 - 17) - ^ «A 

188 
Therefore the surface generated by both loops is 2 S = — — -kcl 2 . 

135 



264 INTEGRAL CALCULUS §179 

179. Exercises. 

1. Find the volume and surface generated by revolving about the 
initial line the cardioid 

p = 2o(l — cos0). 

2. Find the surface generated by revolving the lemniscate 



(1) about the initial line, (2) about a perpendicular to the initial line 
through the pole. 

3. A loop of the curve p = a sin 2 is revolved about the initial line 
and about a perpendicular to this line through the pole. Find the volumes 
generated. 

4. Find the volume generated by revolving a loop of the curve of 
exercise 3 about an axis of symmetry. 

5. Find the volume generated by a loop of the curve of exercise 3 when 
revolved about a perpendicular to an axis of symmetry through the pole. 

6. Find the volume and surface generated by revolving the curve 
p = a sin \ d about the initial line. 



/" 



BOOK III 

INTRODUCTION TO ANALYTIC GEOMETRY OF 
THREE DIMENSIONS 



CHAPTER XXV 



THE POINT, THE PLANE, AND THE SURFACE IN SPACE 

180. Rectangular Space Coordinates. In this and the follow- 
ing chapter are presented a few of the elementary principles of 
Analytic Geometry of Three Dimensions. 

Consider three mutually perpendicular planes whose common 
point is 0, and which intersect in pairs in the three (mutually 
perpendicular) lines X'OX, Y'OY, 
and Z'OZ. The planes are termed 
coordinate planes and the lines co- 
ordinate axes. is termed the ori- 
gin. X'OX is termed the x-axis, 
Y'OY the y-axis, and Z'OZ the ?.'.... 
z-axis. The plane which contains 
both the z-axis and the y-axis, viz., 
the plane XYX'Y', is termed the / 
xy-p\sme, and similarly the other 

coordinate planes are termed the 2/2-plane and the rcz-plane. The 
position of a point in space is determined by three coordinates 
which are the perpendiculars let fall from the point upon the 
coordinate planes. Thus, in the figure the coordinates of P are 
PA, PB, PC. These coordinates are denoted by x, y, z, these 
letters being so chosen that the ^-coordinate shall be parallel to 

265 



266 



GEOMETRY OF THREE DIMENSIONS 



§181 



the z-axis, X'OX, and perpendicular to the yz-jA&ne, the y-coov- 
dinate parallel to the ?/-axis and perpendicular to the zz-plane, 
and the z-coordinate parallel to the 2-axis and perpendicular to 
the #2/-plane. As in plane geometry, the coordinate axes have 
each a + and a — part. OX, OY, and OZ in the figure represent 
the positive parts, while OX' , OY', OZ' are the negative parts. 
The coordinate planes are then conceived to have each a + 

and a — side, the _ j side being that on which lies the _ J 

part of the axis perpendicular to this plane. And accordingly a 
coordinate of a point is + or — according as it lies on the + or 
— side of its plane, i.e., of the plane to which it is perpendicular. 
The three coordinate planes divide the space about the origin into 
eight regions termed octants, and these octants are characterized 
by the signs of the coordinates of the points in them. For example, 
naming the coordinates of a point always in the order x, y, z, it is 
readily seen that the signs of the coordinates of all points in the 
octant XYZO are +,+,+, in the octant XY'ZO are +, — , +, 
in the octant X'YZ'O are —,+,—, etc., etc. 

181. Relations between Two Points. If r be the distance of 
the point (x, y, z) from the origin (see figure in Art. 180), it is 
plain that 

(1) r = Vx 2 + y 2 + z\ 

Further, let Pi and P 2 of the accompanying figure be two 

points with the coordinates 



% 



z 




X 


\m 


/ 








Y' 


\ 












X 


i 




x' 








X 














/ 


Z 




TttN 


rN^ 


/ 








(zi,2/i,zi) and (#2,2/2,22), and let 
r be the distance between 
them. It is evident from the 
figure that the dimensions of 
the parallelopiped of which r 
is the diagonal are x 2 — x h 
yi — yi, an d 21 — 22, and that 
consequently 



(2) 



>~= Vixi-x^y+iyi-y^y+i^-z^. 



182 



THE POINT, THE PLANE, AND THE SURFACE 267 



Again, the coordinates of Q which divides the segment P1P2 in 
the ratio m : n are found without much difficulty to be 



^ mx 2 + nxi 
W m + n > 


my 2 + nyi 

m + n ' 


raz 2 + w«i 
m + n 


m 
or, writing X = — , 






, .v Xi + X^ 2 


Vi + X2/2 


Zi + XZ2 



1 + X ' 1-f-X ' 1 + x 

In obtaining the formulae (1) . . . (4) we have placed both 
points in the first octant : it can be shown that the formulae hold 
for points in any octants. 

182. Some Simple Loci. The equation x = is the equation 
of the 2/z-plane, for it is satisfied by all the points in this plane and 
by no others. Similarly, y = is the equation of the zz-plane, 
and z = of the xy-plane. 

Again, the equation x = a represents a plane parallel to the 
2/z-plane and at a distance a from it. Similarly, y = b, and z = c 
represent planes parallel to the xz-plane and the xij-plsme respec- 
tively. 

Every equation in two of the variables x, y, z is the equation 
of a curve in one of the coordinate planes, but from the standpoint 
of solid geometry such an equation is 
more properly regarded as the equa- 
tion of a right cylinder whose genera- 
tors are perpendicular to the plane of 
the two variables in the equation. 
Consider, for example, the equation 

x z + y 2 = r 2 . 

This represents a circle in the xy- 

plane. This equation is satisfied, 

however, not only by every point in 

this circle, but also by every point in every line through the 

circumference of this circle and perpendicular to the xy-p\a,ne. 

In other words, x 2 + y 2 = r 2 'is the equation of a right circular 





1 

1 
1 
1 

1 
1 

1 


! z 


y/\ 

/ 1 




X' 








— 7T\ 
/ ^^v^ 1 


X 




Y 


z 


i 1 

| | 

1 
! ! 





268 GEOMETRY OF THREE DIMENSIONS §183 

cylinder whose radius is r and whose axis is ZOZ' '. Similarly, 
y 2 + z 2 = r 2 and z 2 + x 2 = r 2 are equations of right circular 
cylinders of radii r whose axes are XOX f and YOY' respectively. 
Again, (x — a) 2 + (y — b) 2 = r 2 represents a right circular cylinder 
of radius r whose axis is perpendicular to the xy-plsme and pierces 
that plane at the point (a, b, 0). 

x 2 y 2 
The equation — + ^ = 1 represents a right elliptic cylinder 

Qj 

whose axis coincides with ZOZ f . 

The equation y 3 — 3 ayz + z 3 = represents the folium of Des- 
cartes in the yz-pl&ne (Art. 89). From 
. , the standpoint of three dimensional 

L^Z geometry, however, this equation repre- 

' sents the right cylinder formed by erect- 
j n g perpendiculars to the yz-plsme at 

/every point of the folium. 
From the foregoing examples it is 
obvious that every equation in two vari- 
ables represents not only a curve in one of the coordinate planes, 
but also the cylinder formed by erecting perpendiculars to this 
plane at each point of the curve. 

When the equation in two variables is linear, the cylinder 
becomes a plane. Thus ax + by + c = represents not only a 
right line in the ^-plane, but also the plane formed by erecting 
perpendiculars to the xy-plsnae at every point of this line. Sim- 
ilarly, ay + bz + c = and ax + bz + c = represent planes 
parallel to the a>axis and to the 2/-axis respectively. 

183. The Plane in Space. The reader will have to accept 
without proof the statement that an equation in x, y, z represents 
some kind of a surface in space, just as he accepts without proof 
the statement that an equation in two variables represents some 
plane curve. 

We shall prove that every equation of the form 
(a) ax + by + cz + d = 

represents a plane in space. 



183 



THE POINT, THE PLANE, AND THE SURFACE 269 



To show this, let (x h y h Zi) and (x 2 , y 2 , z 2 ) be any two points of 
the surface (a). Then 

axi + byi + czi + d = 0, and ax 2 + by 2 + cz 2 + d = 0. 

We multiply the last equation by X, add the product to the first 
equation, and divide the sum by 1 + X. There results 



(b) 



fl g i + ^ + 6 yi + ^» + c gL±^? + d = . 



1+X ' 1 +X 
This equation asserts that the point 
xi + \x 2 2/1 + X2/2 



(X) 



1+X 



Zi + X22 



1+X 



1+X 1+X 

lies upon the surface (a) for all values of X. But (X) is a point in 
the line joining (xi, y i} 21) and (x 2 , y 2 , z 2 ) (Art. 181). Hence this 
line lies wholly in the surface (a). Consequently the surface (a) 
is such that the line joining any two points of it lies wholly in the 
surface, and such a surface is a plane. Q. E. D. 

The line of intersection of a surface with a coordinate plane 
is termed the trace of the surface in that coordinate plane, and 
the equation of the trace is obtained by putting x = 0, or y = 0, 
or z = 0, as the case may be, in the equation of the surface. For 
example, the trace of the surface 

2x 2 -Zxy -y 2 - 4z 2 = 5 

in the ^-plane is obtained by putting y = in the equation of 
the surface, and is 

2x 2 -±z 2 = 5. 
The traces of the plane 
(a) ax + by + cz + d = 
are right lines, and their equa- 
tions in the x?/-plane, the yz- 
plane, and the zx-plane are 
respectively Y ' zX 

ax + by + d = 0, by + cz. + d = 0, ax + cz + d = 0. 
A plane is indicated in a figure by drawing its traces. 




270 GEOMETRY OF THREE DIMENSIONS §184 

If in the equation of the plane (a) we set y = 0, z — 0, we get 

ax + d = 0, whence x = , and this is the intercept of (a) on 

a 

the x-axis. Similarly, — ■=■ and — - are the intercepts of (a) on the 

u c 

y-Sixis and the 2-axis. Equation (a) may be thrown into the form 





_d^ d^ d l ' 

a b c 


and on placing — 
the form 

(i) 


d d , d 
ffli 6i A ' 



= ci, equation (a) takes 



which is the intercept form of the equation of a plane. 

A right line in space cannot be given by a single equation. But 
the equation of any two planes that contain the line are regarded as 
the equations of the line. 

184. Exercises. 

1. Find the mid-point and also the points of trisection of the line 
which joins the points ( — 2, 1, 3) and (3, —5, —4). 

2. Find the points where the line 

3 x + 2 y - 4 z + 7 = 
x-y + 3z-l = 

pierces the coordinate planes. Draw a figure showing these points an i 
the traces of the planes. 

3. Find the vertices and the lengths of the edges of the tetrahedron 
whose faces are 

Sy - z- 1 = 

5x-y + 2z-13 = 

y + 3z- 7 = 

x + y + z— 2 = 

4. Find the equation of the plane determined by the three points 

(1, -1,1), (2,0,1), (-1,2,0). 

5. Prove that S + \S' = is the equation of a surface passing through 
the line or curve of intersection of the two surfaces S = 0, £'= 0. 



x 



§185 THE POINT, THE PLANE, AND THE SURFACE 271 

6. Find the equation of the plane determined by the point (1, 1, —1) 
and the line 

2x -?/4-32-4 = 
x + 5y — z — 1 = 

7. Find the equations of the planes through the line of exercise 6, 
each perpendicular to a coordinate plane. Draw the traces of these 
planes. 

8. Do the two following lines have a point of intersection? 

x + y + z - 1 = and x - y + z + l = 
x — y — s + l = x -\- y — 2 + 1 = 

9. Show that the intersection of the two cylinders 

x 2 + y 1 = r 2 and y 2 + z 2 = r 2 

consists of two plane curves. 

10. Write the equations of the planes tangent to the cylinders of 
exercise 9 at the point (x h y h Si) on each. 

11. Prove that the lines joining the mid-points of the opposite edges 
of a tetrahedron bisect one another. Find the coordinates of this point 
in the tetrahedron of exercise 3. 

185. Quadric Surfaces. As has already been stated, every 
equation in three variables represents a surface. To determine 
the form of the surface from its equation is usually a very difficult 
problem, but, in the case of some of the simpler equations, some 
very elementary considerations may enable us to get a fairly 
good idea of the shape of the surface represented. The surface 
represented by an equation of the second degree in x, y, z is termed 
a quadric surface. Some special cases of quadric surfaces follow. 

1. The Sphere, 

x 2 + y 2 + z 2 = r 2 . 

This is the equation of the surface of a sphere. For it asserts 
that every point on the surface is at the constant distance r from 
the origin (Art. 181). The equation of the sphere whose center 
is at (a, /3, 7) and whose radius is r is 

(x - aY + (y - /3) 2 + O - 7 ) 2 = r\ 

2. The Ellipsoid, 

n 2 "T" 7,2 "T ^2 - 1 - 



272 



GEOMETRY OF THREE DIMENSIONS 



185 



Setting x, y, z successively equal to 0, we get as the equations 
of the traces of the surface in the coordinate planes 



+ ~o=h 



C 2 



+ *_ = 1 



and the traces are seen to be ellipses whose axes coincide with the 

axes of coordinates and whose centers 
are at the origin. 

Writing the given equation in the 
form 

^ 4- yl = 1 _ z l 

n 2 ~T~ 7>2 L „2' 




we note that, because the first mem- 
ber is the sum of two squares, the 
second member can never be negative for real values of x, 
y, and z, and that therefore | z \ = \c\.* Hence the surface lies 
wholly between the two planes z = — c and z= +c. In a simi- 
lar manner it can be shown that the surface lies between the 
planes y = —b and y =-\-b, and between the planes x = — a and 
x = +o. Hence the surface lies entirely within the rectangular 
parallelopiped bounded by these six planes. We shall next deter- 
mine the curves of section of the given surface by planes parallel 
to the coordinate planes. Consider first the plane x = k. We 
substitute k for x in the given equation, and get, after reduction, 



(a) 



y 



b 2 l -. 



+ 



<'-5) 



1. 



This may be regarded as the equation of the right cylinder which 
projects upon the yz-plane the curve of section S, made with 
the surface by the plane x = k, or, as the equation of the projec- 
tion, S f , of this curve of section upon the 2/2-plane, or, as the equa- 
tion of S itself. This curve of section S is, therefore, an ellipse 

* I a I is read " absolute value of a," and denotes the numerical magnitude of a 
without regard to sign. For example, | — 5 | =5, and I +5 | =5. The expres- 
sion J z I := I c ] is therefore equivalent to the longer expression —c=z%-\-c. 



§185 THE POINT, THE PLANE, AND THE SURFACE 273 



whose center is in the z-axis, whose axes lie in the zz-plane and 

/ k 2 
the z?/-plane, and whose semi-axes have the lengths b\/l -^ 

When | k | < | a | this ellipse is real, and when | k \ > \ a \ 



04: 




it is imaginary. Therefore every section of the ellipsoid by a 
plane parallel to the 2/0-plane 
is an ellipse, real or imagi- 
nary. As k increases from 
to a, or decreases from to 
— a, the cutting plane moves — 
away from the origin, and the 
ellipse S grows smaller and 
smaller, until, when a;=±c, 
it has shrunk to a point of the 
z-axis. The solid inclosed by the surface may therefore be 
regarded as made up of a series of elliptic slices strung on the 
x-axis. Exactly similar arguments show that every section of 
the ellipsoid by a plane parallel to the zz-plane, or parallel to the 
xy-pl&ne, is an ellipse, real or imaginary, and that the solid may 
be regarded as made up of a series of elliptic slices strung on 
the i/-axis, or as elliptic slices strung on the 2-axis. We have now 
a good idea of the shape of the surface. It can be proved that 
every plane cuts the surface in an ellipse, but we shall not give 
the proof. The surface (and the solid bounded by the surface) is 
termed an ellipsoid. When two of the quantities a, b, c are equal, 
one set of ellipses becomes circles, and the ellipsoid becomes an 
ellipsoid of revolution (Art. 174). Hence the equation of an ellip- 
soid of revolution has one of the following forms: 

m 2 x 2 -\-y 2 -\-z 2 =r 2 , x 2 +m 2 y 2 -\-z 2 =r 2 , x 2 -\-y 2 -\-m 2 z 2 =r 2 . 

When a = b = c, the ellipsoid is a sphere of radius a. 
3. The Hyperboloid of One Sheet, 



x _u y _ _ 

a 2 + b 2 ~ c 2 



274 GEOMETRY OF THREE DIMENSIONS §185 

It is easily seen that the trace in the xy-plsme is an ellipse, and 
that the traces in the zz-plane and the yz-plsme are hyperbolas. 
The z-axis is the conjugate axis of each of these hyperbolas, and 
consequently the four foci lie in the xy-pl&ne. 

Setting z = k in the equation of the surface, we get, after easy 
reductions, 

X 2 y2 _ 

~> / 7 o\ — J-) 



which is the equation of the right cylinder which projects upon 
the £?/-plane the curve of section of the 
given surface by the plane z = k, or is the 
equation of this projection, or the equa- 
tion of the curve of section itself. Hence 
every section made by a plane z = k is 
an ellipse, and this ellipse becomes larger 
as | k | increases. The solid bounded by 
the surface may be conceived as made up of a series of ellipses 
strung upon the z-axis. 

On writing y = k in the equation of the surface and transform- 
ing, there results 




(-» '(>-» 



The section is an hyperbola. When | k \ < \ 
and foci are in the £?/-plane, and its conjugate axis is in the 
2/z-plane. When \k\ > \b\, the transverse axis and foci are in 
the 2/z-plane and the conjugate axis is in the :n/-plane. When 
k = b, the section is a pair of right lines intersecting in the y-axis. 
Similar remarks apply to the sections made by the planes 
x = k. Let the student give the complete discussion of these 
sections. 

4. The Hyperboloid of Two Sheets, 

x 2 _ y 2 _ z 2 

a 2 b 2 c 2 



§185 THE POINT, THE PLANE, AND THE SURFACE 275 



Writing this in the form 



t. _l z l = *L. 

i' ' -« -,2 



1: 



b 2 C* a 1 

it is seen that when | x \ < \a\,y and z cannot both be real, which 

means that the surface has no real points between the planes 

x = — a and x = +a. The traces in 

the xy-pl&ne and the zz-plane are 

hyperbolas with their foci in the 

o;-axis, while the trace in the yz- 

plane is of course imaginary. It can 

be shown without difficulty that the 

planes y = k,z = k, cut the surface in hyperbolas, while the planes 

x = k, (| k | > | a J), cut it in ellipses. The solid bounded by the 

surface may be conceived as made up of ellipses strung on the 

x-axis. 

5. The Elliptic Paraboloid, 







When z is — , x and y cannot both be real, which means that the 
surface has no real points below the xy '-plane. It is readily found 
that the traces in the xy-plsme and xz-p\sme are parabolas turned 
upward, while the trace in the xy-ipl&ne is the origin. The planes 
z = k cut the surface in ellipses which increase in size as k in- 
creases. The surface may be conceived as built up of ellipses 
strung on the 2-axis. Let the student complete the discussion 
and draw the figure of this surface. 

6. The Elliptic Wedge, 

x 2 z 2 + a 2 y 2 = b 2 x 2 . 
Writing the equation in the form 

ay = ±x Vb 2 — z 2 , 

it appears that the surface lies entirely between the planes z = — b, 
z = +6. When z = d=6, y = 0, and therefore the trace in the 
zz-plane consists of two parallel lines z=—b, z = -\-b. Every 
plane parallel to the xy-plane, z = ±k, cuts the surface in a pair 



276 



GEOMETRY OF THREE DIMENSIONS 



186 



of right lines, ay = zLx Vb 2 — k 2 , the xz-plane bisecting the angle 
between them. When x = dbk, 

k 2 z 2 + a 2 y 2 = b 2 k 2 or ^ + £- = 1. 



Hence the planes parallel to the 1/2-plane cut the surface in ellipses. 

When \k\ < \a\, these ellipses have their foci in the zz-plane, and 

when | k \ > | a |, the foci are in 
the rn/-plane. One semi-axis of 
each of these ellipses is b. When 
| k | = | a |, the section is a circle 
of radius b. The surface is sym- 
metrical as to each of the coordi- 
nate planes, and intersects itself 

in the 2-axis. Only the right-hand portion of the surface is shown 

in the figure. This of course is not a quadric surface : it is a 

quartic surface. 

186. Exercises. 

1. Show that x 2 -f y 2 -f z 2 + Ax + By + Cz + D - represents a 
sphere, and find its radius and the coordinates of its center. 
Determine the shapes of the following surfaces: 




y. +.* = i 

6 2 c 2 



3. x 2 + y 2 - a 2 z 2 = 6 2 ; 



4. -* + * 



a* 



5. aV 



fi — 9% 



b 2 ; 



6. - 2 + - 2 =22/; 
a 2 c 2 

7. x 2 + y 2 = 2a 2 z; 



.0 I 7.0 ' -9 ■"■• 



f 2 /7 2 ?/ 2 



a 2 ?/ 2 + z 2 = 6 2 . 



= 1; - 



+ -9=1. 



— T 2 - 



i/ 2 + a 2 z 2 = b 2 . 
- = 2x. 



y 2 + z 2 = 2 a 2 z. 
a; 2 z 2 

^-3 = 2 2/; 



2/ z 



= 2x. 



The surfaces of exercise 8 are termed hyperbolic paraboloids. 
9. xz = y. 10. z 2 + y 2 = (3 - j/)*. 



11. 



,2 ?4 

+ 



1. 



12. £3 -f- 2/3 -f- £3 = 3. 



§186 THE POINT, THE PLANE, AND THE SURFACE 277 

13. z = cos VtfTy 2 : x 2 + y 2 = cos z. 14. z = sm V * 2 + J / 2 . 

V x 2 + y 2 

15. What are the mutual relations of the following sphere and cylinders? 
x 2 + y 2 + z 2 = 2, 2 r* + ?/ 2 = 3, y 2 -f 2 f = 1, x 2 - z 2 = 1. 

16. A homogeneous equation in three variables is one in which all the 
terms are of the same degree in all the variables. For example, 

x 2 + 3 xy — z 2 + yz = and x z -\- y z -\- 2> y 2 z — b xyz = 

are homogeneous in x, y, z. 

Show that every homogeneous equation in x, y, z represents a cone 
with vertex at the origin. 

17. Determine the shapes of the following surfaces: 

- 2 + h~% = 0; xy = z 2 ; xy + yz + zx = 0; z z = x 2 y. 
a b c 

18. Given the tetrahedron whose vertices are 

(x 1} y h z x ), (x 2 , y 2 , z 2 ), (x 3 , y 3 , z s ), (x i} y i} 2 4 ); 

prove that the lines joining the mid-points of opposite edges meet in a 
point, and find the coordinates of this point. 

19. Given the triangle whose vertices are 

(si,yi, Si), (x 2 ,y 2 , z 2 ), (x z ,y 3 ,z 3 ); 

find the coordinates of the intersection of its medians. 

20. In the tetrahedron of exercise 18 prove that the lines joining the 
vertices with the point of intersection of the medians of the opposite faces 
meet in a point, and find its coordinates. 



CHAPTER XXVI 



M' 


P>-£~ 




\Q" 


M' 


M 


F 




L- 


M 




P' 




y. 


\^ 



THE RIGHT LINE IN SPACE 

187. Projection. The angle between two lines L and M lying 
in different planes is defined to be the angle which either line 
makes with a line drawn through any point of it parallel to the 
other line. Let L and M be any two lines, and let M' be drawn 

parallel to M through a point 
P of L. Then d, the angle 
between L and M', is by defi- 
nition the angle between L 
and M. 

Let PQ be a segment of L. 
Through P and Q let planes 
(or lines) be passed perpendicular to M, and cutting M in the points 
P' and Q' '. P' and Q' are defined to be the projections of the points 
P and Q upon M, and the segment P'Q' is defined to be the pro- 
jection of the segment PQ upon M* 
It follows from the figure that 

(a) P'Q' = PQ" = PQ cos 0, 

or, in words : 

A. The projection upon any line of a line-segment PQ is the 
product of PQ by the cosine of the angle between the lines. 

These definitions and principle A hold also when L and M lie 
in the same plane. 

* The projection defined in the text is orthogonal projection. If the planes 
through P and Q were not perpendicular to M, the segment P'Q' would be 
the projection of PQ in a wider sense. It is when the planes are perpendicular 
to M that the projection is orthogonal. When the term projection is used 
in this book it always means orthogonal projection. 

278 



§188 



THE RIGHT LINE IN SPACE 



279 




Let AB, BC, CD, . . . PQ be the segments of a broken line, 
these segments lying in the same or in different planes, and let 
A', B', C, . . . P', Q' be the 
projections upon the line L of 
the points A, B, C, . . . P, Q. 
Then A'B', B'C, . . . P'Q! are 
in magnitude and sign the pro- 
jections upon L of the successive 
segments of the broken line and 
A'Q' is the projection of AQ upon L. Now it is evident that 

(b) A'B' + B'C + . . . + P'Q' = A'Q', 
or 

B. The algebraic sum of the projections upon any line L of the 
segments of a broken line is equal to the projections upon L of the line 
joining the extremities of the broken line. 

The broken line and the line QA form a closed polygon, which 
is termed "twisted" when the points A, B, C, . . . P, Q do not 
all lie in the same plane. Moreover, since Q'A' = —A'Q', equa- 
tion (b) may be written in the form 

(b') A'B' + B'C + . . . + P'Q' + Q'A' = 0. 

Expressing this in words, we have as the equivalent of B : 

B'. The algebraic sum of the projections upon any line of the sides 

of a closed polygon, whether plane or twisted, is zero. 

If a, |8, . . . 6, o) be the angles which the segments AB, BC, . . . 

PQ, AQ make with L, it follows from (a) and (b) that 

(c) AB cos a + BC cos jS + . . . + PQ cos 6 = AQ cos w. 

188. Direction Cosines. The direction angles of a line are the 
angles which the positive direction of the line makes with the posi- 
tive directions of the axes of coordinates. If the line does not 
pass through the origin, we draw through the origin a line L, 
parallel to the given line, and then, by our definition of the angle 
between two lines, the direction angles of L are the same as the 
direction angles of the given line. 



280 



GEOMETRY OF THREE DIMENSIONS 



188 



We represent these direction angles by a, /3, 7, as shown in the 
figure. Of course we may choose either direction along a line to 

be the positive direction. In 
the figure, L is so situated and 
the positive direction (indi- 
cated by the arrow) is so 
chosen that all the direction 
angles are acute. Now a little 
reflection will convince the 
reader that, in whatever oc- 
tants L may lie, we can always 



-v 



z 








¥ L> 


/ 






p 


JT L 




^\s& 


1\X^" 










z 







^A& 


X 




X 




'I* 




/ 





so choose the positive direction of L that at least two of its direc- 
tion angles shall be acute, and none shall be greater than ir. 

Thfr direction cosines of a line are the cosines of its direction 
angles, and from what has just been said it follows that at least 
two of the direction cosines may be taken positive. 

Let P with the coordinates x, y, z be any point of L, and let 
OP = r. Then from the figure 

x = r cos a, y = r cos (3, z = r cos y, 
and 

r 2 cos 2 a + r 2 cos 2 (3 + r 2 cos 2 y = x 2 + y 2 + z 2 = r 2 , 
whence 
(I) cos 2 a + cos 2 jS + cos 2 7 = 1, 

which is an important relation among the direction cosines of a 
line. For sake of brevity we set 

I = cos a,m = cos /3, n = cos 7, 
so that (I) takes the form 
(I) I 2 + m 2 + n 2 = 1. 

Observe that any three real numbers which satisfy (I), that is, the 
sum of whose squares is unity, are the direction cosines of some 
line. Further, any three real numbers whatever, as a, b, c, are 
proportional to the direction cosines of some line. For, on divid- 
ing these numbers by Va 2 + b 2 + c 2 , we have 
a b c 



Va 2 + b 2 + c 2 Va 2 + b 2 + c 2 Va 2 + b 2 + 



§189 



THE RIGHT LINE IN SPACE 



281 



three numbers which satisfy (I), and may therefore serve as the 
direction cosines of a line. 

Let a, |8, 7 and a , /3', y' be the direction angles, and I, m, n 
and V ', m', n' the direction cosines of any two lines in space, and 
let be the angle between the lines. Through the origin we draw 
L and L' parallel to the given lines (figure of page 280) . 

Let P with the coordinates x, y, z be any point of L. By prin- 
ciple B of Art. 187, the sum of the projections upon L' of x, y, 
z is equal to the projection upon L' of OP = r. Hence by (c), 

Art. 187, 

Vx + m'y + n'z = r cos 0. 

But by (a), Art. 187, 

x = rl, y = rm, z = rn, 
and therefore 

(II) cos 9 = IV + mm' + nn\ 

a formula which expresses the cosine of the angle between two 
lines in terms of their direction cosines. When the lines are per- 
pendicular, cos 6 = 0, and 
(Hi) IV + mm' + nn' = 0. 

If the lines are parallel they have the same direction cosines, that is, 
(II 2) 1 = 1', m = m', n = n'. 

Problem. Find the direction cosines of the line which joins the points 
(xi, y h 21) and (x 2 , y 2 , z 2 ). 



189. Parametric Equations 
of the Line. A line is com- 
pletely determined by its direc- 
tion and a point upon it. Let L 
be a line determined by the di- 
rection cosines I, m, n, and the 
fixed point (x , y , z ). Let 
(x, y, z) be any other point of 
the line, and let r be its distance 
from (xq, yo, z ) . From the figure 
we have 

x— Xq= r cos a = lr, y — y Q = r cos /3 



J»o.l/o»*c 





/ Kt 




A 








\r 




x-x\ 










X 
















/ 





mr, z — Zq = r cos 7 = nr, 



282 



GEOMETRY OF THREE DIMENSIONS 



§190 



whence 

(III) x = x + Ir, y = yo J r mr, z = Zq-\- nr. 

Now if r be variable, so also are x, y, and z, and as r varies from 
— oo to +oo, the point (x, y, z) traces the line. Equations (III) 
are termed the parametric equations of the line, r being the variable 
parameter. 

190. The Normal Equation of the Plane. Let p be the length 
and a, (3, y the direction angles of the perpendicular (the normal), 

OQ, let fall from the origin 
upon a given plane. Let P 
with the coordinates x, y, z be 
any point of the plane. Now 
pis the projection uponOQ of 
ur -X the line OP. Hence by B or 
(c) of Art. 187 we have 

x cos a + y cos /3 + z cos y = p 
or 

(IV) Ix -j- my + nz = p 

and this is the normal equation of the plane. 

The equation 

Ax + By + Cz + D = 

can be thrown into the normal form by dividing by the square 
root of the sum of the squares of the coefficients of x, y, and z. 
Thus, if R = A 2 + B 2 + C 2 , then 




(V) 



A , B , C D 

—==. x + — -7= y H 7= 2 = 7= 

Vfl VR y Vr Vr 



is the normal equation of the plane. Comparing (V) with (IV) , 



we see that 

to the plane, and 



A, JL, JL 

Vr Vr Vr 

-D 



are the direction cosines of the normal 



Vr 



is the length of this normal, the radical 



being given the sign contrary to that of D, so that 



-D 

Vr 



shall be 



§191 THE RIGHT LINE IN SPACE 283 

positive. From (V) it follows that in the equation of any plane the 
coefficients of the variables are proportional to the direction cosines 
of the normal to the plane. 

The equations of a line through the point (xq, Vq, z ) normal 
(perpendicular) to the plane Ax + By + Cz + D = are there- 
fore 
(VI*) x = x + — ,= r, y = 2/o + —?= r, z = z + —== r, 

where r is the variable parameter. 

f 
Now we may write r in place of —=. , and the equations of the 

VR 

normal line take the simpler form 

(VI) x = Xq + Ar, y = y -\- Br, z = z + Cr. 
The angle between two planes 

Ax + By + Cz + D = and A'x + B'y + C'z + D' = 
is equal to the angle between their normals, and if 6 be this angle, 
then by (II) 

firm a AA'+BB'+CC 

(VII) cos 6 = 



VA 2 + B 2 + C 2 VA' 2 + B' 2 + C' 2 

Again, the angle between the plane Ax -\- By -\- Cz -{- D = 0, and 
the line whose direction cosines are I, m, n is the complement of 
the angle between the line and the normal to the plane. There- 
fore by (II) 

(VIII) sin fl = l A+mB + ng m 

,\ / A 2 + B 2 + C 2 

It has already been explained, Art. 183, end, that the equations 
of two planes that contain a line are termed the equations of the 
line. 

191. Exercises. 

1. What are the conditions that the planes 

Ax + By + Cz + D = and A'x + B'y + C'z + D' = 
shall be perpendicular to each other? parallel? 

2. What is the condition that the plane 

Ax + By + Cz + D = 



284 GEOMETRY OF THREE DIMENSIONS §191 

and the line 

x = x + Ir 
y = y + mr 

z = Zo + nr 

shall be perpendicular? parallel? 

3. A line through the point (xi, yi, Zi) is perpendicular to the plane 

Ax + By + Cz + D = 0. 

Find its equation, and the coordinates of the point where it pierces the 
plane. 

4. ^ind the equation of the plane through the point {x\, y h Zi) per- 
pendicular to the line of exercise 2. Find the point of intersection of the 
plane and the line. 

5. Find the equation of the plane through the points (1, 1, 1) and 
(1, — 1, 2) perpendicular to the plane 

x + y-z + l = 0. 

6. Find the equation of the plane through the point (xi, y h Zi) parallel 
to each of the two lines whose direction cosines are h, mi, n h and l 2) m 2 , n 2 . 

7. What are the direction cosines of' the line through the points 
(1, — 1, 1) and (3, 3, 5)? Write the parametric equations of this line, 
and find the angle which it makes with the plane 

x + 2y -z + l = 0. 

8. Find the direction cosines of the line 

2x + y -3z + 1 = 
x-y + z + 2 = 0. 

Hint. The line is perpendicular to the normals to the two given planes. 

9. Find the equation of the plane through the origin perpendicular 
to the line of exercise 8. 

10. Find the equation of the plane through the origin making an angle 
of 30 degrees with each of the lines 

x = xi + r x = x- 2 — r 

V =Vi — r and y = y 2 + r 

z = Zi-\- r z = z 2 + r 

11. Find the equation of the line through the point (x Q , y , Zo) 

! P araUel dlCUlar 1 to the line ^ oining the P oints (^> Vh Z J and (x 2 , y 2 , z 2 ). 

12. Find the angle between the lines 

2x-y + z-3 = and x + y-2z + 5 = 
x + 3y - 2z + 1 = 2x + y + 3z- 1 = 



§191 THE RIGHT LINE IN SPACE 285 

13. Do the lines of exercise 12 have a point in common? 

14. Let (x , yo, z ) be the foot of the perpendicular let fall from 
(xi, 2/1, 2i) upon the plane 

Ax + By + Cz + D = 0. 

Show that the length of this perpendicular is 

Axi + %i + Czi + D 

\/A 2 + 5 2 + C 2 

fl"t7ii. Use equations (VI*) of the preceding article. 



CHAPTER XXVII 



VOLUMES OF SOLIDS. AREAS OF CYLINDRICAL 
SURFACES 

192. Volumes of Solids. The method of Art. 175, for finding 
the volume of a solid of revolution, is applicable to finding the 
volume of any solid. This method consists in regarding the 
volume as the limit of the sum of an infinite number of right 
cylinders of infinitesimal altitudes, and in calculating this limit 
by integration. If A (x) be the area of the base of such a cylinder 
at distance x from the yz-pl&ne, and if dx be its altitude, then 

X = b r> h 

V = lim Z,A(x) dx = J A{x) dx. 



1. To find the volume of the ellipsoid, 



+ 



y- 



+? 



l. 



Conceive the ellipsoid to be cut into thin slices by planes parallel 
to the i/z-plane. Each of these planes cuts the-slirface in an ellipse. 

On each of these ellipses as base 
construct a right cylinder whose 
other base lies in the adjoining 
plane. The volume of the ellip- 
soid is the limit of the sum of 
the volumes of these cylinders 
as their number increases with- 
out limit. This is fairly obvious 
from the figure, and the statement can be proved rigorously by a 
method similar to that used in Arts. 175 and 176. We do not 
give the proof. Let the bases of one such cylinder lie in the planes 

286 




/" 



§192 



VOLUMES OF SOLIDS 



287 



x = Xi and x = X\ + dxi. Setting x\ for x in the equation of the 
ellipsoid, we get, after simple transformation, 



r 



+ 



= 1. 



This is the equation of that base of our cylinder which lies in the 
plane x = X\. The area of this base is irbcl 1 —J, and there- 
fore the volume of the cylinder in question is irbcl 1 jjdxi. 

Dropping subscripts, we have for the volume of the ellipsoid 

v= Thc J-a \ l " a*r = Tbc l x ~ 3^J_ a = r abc - 

The same result would of course have been obtained by cutting 

the ellipsoid into slices parallel to either of the other coordinate 

planes. Note that when two of the quantities a, b, c are equal, the 

foregoing formula gives the volume of the ellipsoid of revolution, 

and when a = b = c, the formula gives the volume of the sphere. 

x^ v^ 
2. To find the volume cut from the paraboloid — + j- 2 = 2 z by 

the plane z = mx. 

The solid whose volume is required is OAPBC of the figure. 
We conceive the solid to be cut into thin slices by planes parallel 
to the 2/2-plane. These 
planes cut the parabo- 
loid in parabolas, such 
as ACB. On these 
parabolas as bases we 
suppose right cylinders 
to be constructed, and 
the volume required is 
the limit of the sum of 
these cylinders. This 
last statement also ad- 
mits of rigorous proof, which we omit. Consider the cylinder one 
of whose bases, ACBD, lies in the plane x = x h and whose other 




P(2a 2 m,2a 2 m 2 ) 



288 INTEGRAL CALCULUS . §192 

base lies in the plane x = x\ + dx\ (not shown in the figure). 

The area of this parabolic section ACBD is § AB X CD (Art. 160, 

example 2). We must express this area in terms of X\, C is the 

point of intersection of the line (or plane) x = X\ with the trace 

of the paraboloid in the zz-plane. The equation of this trace is 

x 2 ( Xi 2 \ 

-s = 2 z, and C is therefore x h 7—^ ). D is the intersection of 

a 2 \ 2 a 2 / 

x = xi with z = mx, and is the point (x h mx\). Hence 

mt. Xi 2 2a 2 mxi — Xi 2 

CD = mx\ — 77—= = 



2 a 2 2 a 2 

The equation of the parabola ACB is found by putting x = Xi in 
the equation of the paraboloid, and is 

b 2 ~ ZZ a 2 

When z = mx\ we have 

y 2 _ x x 2 2a 2 mx 1 -x 1 2 b ^r^—. 2 

fz = 2 mil =■ = o ? and 2/= - V2 a 2 razi - Z1 2 , 

o 2 a 2 a* a 



whence 



Therefore 



AB = 2 y = — V2 a 2 mx x - x x \ 
a 



Area ACBD = %AB-CD = ^\(2a 2 mx 1 - m 2 )l 

The volume of the cylinder in question is the product of this by 
dx. The coordinates of P. are easily found to be 2 a 2 m, 2 a 2 m 2 , 
and therefore, dropping subscripts, we have for the volume required 
the formula 

*x=2a 2 m 3 

(2 a 2 mx - x 2 Y dx. 

=0 



3a 3 J*=( 



To integrate this, we write 

2 a 2 wx — z 2 = a 4 m 2 — (a 2 m — x) 2 = a 4 ?^ 2 — w 2 , 
where u = a 2 m — x. 



§§193-194 VOLUMES OF SOLIDS 289 

Then 

V=~ (a 4 m 2 -u 2 )Uu. 

O 0, Ju= —a 2 m 

Integrating by tables, we get 

TT a 6 bm* 

V = — -. — x. 
4 

193. Exercises. 

1. Find the volume of that portion of the hyperboloid of one sheet, 

— + t-„ -k = 1. which is cut off by the planes z = 0, z = c. 

a 2 b 2 c 2 

7/2 gl 

2. Find the volume cut from the paraboloid f- + - = 2 x, by the 

b l c 2 

plane x = a. 

3. Find the volume of each wedge cut from the cylinder x 2 + y 2 = r 2 
by the planes z = mx, z = 0. 

Solve in two ways, by cutting the solid into slices: (1) parallel to the 
zz-plane; (2) parallel to the 7/2-plane. 

4. Find the volume of the solid cut from the cylinder x 2 — 2 rx + y 2 = 
by the planes z = mx and 2 = 0. Solve by cutting the solid into slices: 
(1) parallel to the zz-plane; (2) parallel to the 2/2-plane. 

5. Find the volume common to the two circular cylinders x 2 + y 2 = r 2 
and y 2 + z 2 = r 2 . 

6. Find the volume of that portion of the elliptic wedge x 2 z 2 + a 2 y 2 — \b 2 x 2 
which is included between the planes x = and x = a. 

7. Find the volume of the solid bounded by the surfaces x 2 + y 2 = az, 
x 2 + y 2 = 2 ax, and z = 0. 

8. Find the volume cut from the paraboloid y 2 + z 2 = 4 ax by the 
plane z — z = a. 

9. Find the volume inclosed by the surface x^ + y* + z^ — a*. 

x 2 11 2 z^ 
10. Find the volume inclosed by the surface — + tt, + —. = 1. 

a 2 b 2 c 4 

194. The Line Integral. The Area of a Cylindrical Surface. 

Let AB be an arc of the curve y = g(x), and letf(x, y), a function 
of two arguments, be real, single-valued, and continuous along AB. 
Divide the arc, AB, into small intervals or arcs, As, and form the 

B 

sum of products, 2jf(x,y)As. Let the number of small arcs be 
J? 

B 

increased in such a way that each As == 0. Then V f(x, y)As has 



290 



INTEGRAL CALCULUS 



§194 



Z> X 



a limit. This is evident geometrically. For, if at each point of 
division of AS we erect a ^-ordinate whose length is equal to the 

value of fix, y) at that point, 
the upper ends of these 2-ordi- 
nates lie in the surf ace 2 =f(x, y) , 
and the 2-ordinates themselves 
form a portion of a cylindrical 
surface ABB' A', and the area 
of this surface is plainly the 

B 

limit of ^f(x,y)As. 

A 

This limit is defined to be the 
line-integral of f{x, y) along AB, 

and is denoted by J f(x, y) ds, so that 

B nB nB 

\im^f(x, y)As= J f(x, y)ds = J 




zds = area of ABB' A'. 



This line-integral is evaluated by expressing fix, y) in terms of x, 
and ds in terms of x and dx by means of the equation y = g(x), 
and thus reducing the line integral to the form of an ordinary 

integral, I F(x)dx. Such a reduction can also be made the basis 

B 

of a proof that^/fe y)^ s nas a limit. It should be noted that 

A 

the line-integral is built up in precisely the same way as was the 

integral 1 f{x)dx of Art. 169, except that in the line-integral the 

several products, f(x,y)As, are calculated for the points of a 
curve, y = g(x), while in the ordinary integral the products are 
calculated for the points of a straight line, viz., the z-axis. Indeed, 
the ordinary integral may be regarded as a special case- of the 
line-integral. The curve A'B' of the figure is the intersection of 
the surface z = f{x, y) with the cylinder y = g(x). 



195 



AREAS OF CYLINDRICAL SURFACES 



291 



Example. 

A circular cone has its vertex at the origin and its axis in coincidence 
with OZ. Find the surface of that portion of the cylinder (x — a) 2 -\-y 2 = a 2 
which is included between the cone and the 
,:n/-plane. 

The equation of the cone is 

x 2 + y 2 - m 2 z 2 = 0, 
m being the tangent of half the vertical angle. 
From this, 



z — 



Vx 2 + 



Denoting that part of the cylinder which lies 
in the first octant by S, we have 

J.x=2o 1 r x=2a _ 

z ds = - Vx 2 + y 2 ds, 

x =o m J x =o ■ 

the integral to be taken along the semicircle R. Now from the equation 
of the cylinder we get 




= V2 ax — x 2 , ds 



a dx 



Hence 



S = 



f 



V2 ax — x 2 

ix 

1 1 V2 a — x 



■Vx 2 + y 2 = \/2ax. 



VYa r 2a dx 



2 -^^[V2a-=x] 2a =^, 
m L J o m 



195. Exercises. 



1. Through a sphere of radius r el circular hole is bored of radius \ r, 
in such a way that a diameter of the sphere lies in the surface of the hole. 
Find the area of the cylindrical surface of the hole in the sphere. 

2. Find the surface of the cylinder (x — a) 2 + y 2 = a 2 included be- 
tween the £?/-plane and the surface x 2 + y 2 = 2 z. 

3. Find the surface of that part of the cylinder x* + y~s = a* which is 
included in the sphere x 2 + y 2 + z 2 = a 2 . 

4. The axes of two circular cylinders, each of radius a, intersect at 
right angles. Find the surface of each inclosed by the other. 

5. Find the surface of the cylinder x 2 + y 2 = a 2 included between the 
xy-pl&ne and the lower part of the surface of the cylinder y 2 + (z — a) 2 = a 2 . 

6. Find the surface of the cylinder x* + yt = as which is included 
between the z2/-plane and the surface x 2 + y 2 = 2 z. 

7. Find the surface of the cylinder x* + y* = a* which is included 
in the cylinder y* + zs = as. 



292 INTEGRAL CALCULUS §195 

8. Find the surface of the cylinder xa -f ya = a* included between 
the xy-plsme and the cone x 2 + y 2 — m 2 z 2 = 0. 

9. Find the area of that portion of the first arch of the cylinder y = sin x 
included between the xy-plsme and the cylinder z = sin x. 

10. Find the area of that portion of the first arch of the cylinder y = sin x 
which lies in the first octant and is included in the cylinder z = cos x. 



BOOK IV 
FUNCTIONS OF MORE THAN ONE ARGUMENT 



CHAPTER XXVIII 

PARTIAL AND TOTAL DERIVATIVES 

196. Definitions. In Chapter I we learned that a variable 
may be a function of two or more other variables, and examples 
of such functions were there given. We note here, as an additional 
illustration, that any one of the coordinates of a point on a surface 
is a function of the two other coordinates of that point. 

When a function depends upon two or more arguments, these 
arguments may be entirely independent, that is, it may be possible 
to assign to each of them any values whatever, or they may in 
turn be functions of one or more other variables. 

We learned, too, in Chapter I that functions of two arguments, 
of three arguments, etc., are denoted by such symbols as f(x, y), 
F(%,y)', f(x,y,z),<f>(x,y,z)', etc. And just as y= f(x) or F(x,y)=0 
is the equation of a plane curve, so z = f(x, y) or F(x, y, z) — is 
the equation of a surface. This surface is the graph of the func- 
tion z or f(x, y). 

A discontinuity of a function of two variables, f(x, y), may be 
denned geometrically as a point at which the surface which is 
the graph of the function is punctured or torn. If it is torn or 
cut, the severed edges may or may not be sundered. 

The analytical definition which covers every case is as follows: 

V Um f(x + Ax, y + Ay) = f(x , t/ ), 

Ax=0 Ay=0 

293 



294 DIFFERENTIAL CALCULUS §197 

no matter in what way Ax and Ay approach their limits, 0, f(x,y) 
is continuous at (xo, y Q ). If, on the other hand, Ax and Ay can be 
made to approach in such a way that the above equation does not 
hold, then f{x,y) has a discontinuity at (x , y ) . 

The points at which f(x, y) is discontinuous may be discrete or 
they may constitute a curve upon the surface. 

In this book we shall have very little to do with discontinuities 
of functions of more than one argument. 

197. Partial Derivatives. Let z be a function of x and y. 
When y is held fast while x takes an increment Ax, the correspond- 
ing increment of z is termed the partial increment of z as to x, and 

A x z 
is denoted by A x z. Then lim ~- is termed the partial derivative 

ax=o Ax 

dz 
of z as to x, and is denoted by — , or D x z, or z x . Again, when x is 

held fast while y is given the increment Ay, the corresponding 

increment of z is termed the partial increment of z as to y, and is 

A z 
denoted by A y z. Also lim -~ is termed the partial derivative of z 
A y=o &y 

as to y, and is denoted by — , or D y z, or z y . If the functional 

symbol is/(z, y), the partial derivatives are denoted by 

Observe that the expressions — , ^—^- — are not fractions. The 

— and — are single symbols like D x , D y , and the partial derivatives 
ox oy 

are often written —z, -jr-/fo y)> etc. If z is a function of several 
ox ox 

arguments x,y,t, . . . , there are as many partial derivatives as 
there are arguments, and they are denoted by — , — , — , . . . . 



§198 PARTIAL AND TOTAL DERIVATIVES 295 

Examples. 

1. To find the partial derivatives of z when z = x 2 sin y + V log x. 
Differentiating first as though y were constant, and second as though x 
were constant, we have 

— = 2 x sin y + - and — = x 2 cos w + log x. 
dx x dy 

2. To find the partial derivatives of u when u = x 2 + ?/ 2 + z 2 + 2 x?/z. 
Differentiating first as though ?/ and z were constant, next as though z 
and x were constant, and lastly as though x and y were constant, we have 

u x = 2 (x + yz), u y = 2(y + xz), u z = 2 (z + 23/). 

3. To find the partial derivatives of z when x 2 + y 2 + z 2 + 2 xyz = 0. 
Regarding ?/ as constant, we have 

o 1 o dZ . . n dZ n , dZ X + VZ 

2 x + 2 z h 2 vz + 2 xv — = 0, whence — = —*- • 

dx ). dx ' ax z + xy 

Regarding x as constant, we have 

2</ + 2z^+2xz + 2x?/^ =0, whence %.=-U±J*. 
ty ty dy z + xy 

4. Suppose that, in the foregoing example, x is function and y and z 
arguments. We have then 

j. ± o dx 1 o 10 ax . n , ax W + XZ 

z constant, 2x — + 2 v + 2 vz h 2 xz = 0, whence — = — ^— \ 

dy y y ay^ dy x + yz> 

y constant, 2x — -\- 2z -\-2yz \-2xy = 0, whence — = lt_J/ . 

y ' az l * az • . az x + ?/z 

Observe that — • — 5^ — 

dx dy dy 



198. Exercises. 

1 Tr X + V i_ at. x dZ 2 W az 2 X 

1. If z = — —z , show that — = - ^— „, — = - 

x — y dx (x — y) 2 dy (x — yy 



2. If z = tan -1 -, find — and — • 

y dx dy 

3. If u = (x + 2/ + 2) log (x + y + z), show that D x u = Z)„tt = DgW. 

4. If it = x log y -\-y log z + z log x, show that 

du . du . du . . . 

ax dy dz 

5. Find i*. and it- from w = sin -1 - + Vr 2 — 2 . 

9 r 



296 



DIFFERENTIAL CALCULUS 



§199 



6. If xyz = 1, show that 

dz dz n dy dy 

x y — = 0, x-Z — z- 31 

dx dy dx dz 



0, y 2 — = 0. 

* 31/ 32 



/>*2 n /2 ^2 

7. Find the partial derivatives of z from — + 7- — — = 1. 

a 2 b 2 c 2 

8. Find z x , z y , y x , y z , x y , x z from (x + y) 2 + (z + ^) 2 + 2 2?/ = a 2 . 



9. From ?/ = 



z + z 



, find D x ?/, D z y, D x z, D y z, D y x, D z x. 



10. Find all the partial derivatives from tan" 1 - + tan l - 

x u 



1 = 0. 



199. The Tangent Plane and the Normal Line. In the case 
of a function of two arguments there is a simple geometric interpre- 
tation of the partial derivatives. 
Let f(x, y) be such a function; 
then z = f(x, y) is the equation 
A xZi of a surface which is the graph 
± f of the function f(x, y) or z. It is 
obvious that the increments A#i, 
Ai/i, A Xl 2i, A yi 2i are as repre- 
_y sented in the figure. The plane 
y = 1/1, which passes through P 
parallel to the zz-plane, cuts the 
surface in the plane curve PA 
whose equation is z = }{x, yi). 
PR is the tangent to this curve at P, and its slope is obviously 

dZ] 
dx-i 




(i) 



- , where — L = \ — \ . Hence the equation of PR 
1 dxi \_dx_\ x=Xl 

dzt 



IS 



Z=Zy 



Z — Z\ 



dxi 



(x — Xi). 



Similarly, the plane x = X\ cuts the surface in the curve z = f(x h y), 
and PS is the tangent to this curve at P. The slope of PS is 

-r-^ , where -r-^ = — - , and therefore the equation of PS is 
dyi dyi ldy] x = Xl 

V=V\ 
z=z x 

(2) z -*. = ^(y -vx). 



IS 



§199 PARTIAL AND TOTAL DERIVATIVES 297 

We have here the geometrical interpretation of — and — ■ It 

dx dy 

is now a simple matter to determine the equation of the tangent 

plane to the surface at P. This tangent plane cuts the planes 

x — X\ and y = y\ in the tangent lines PS and PR. The equation 

of any plane through P is 

(3) A(x- x,) + B(y - Vl ) + C(z - z x ) = 0. 

The plane y = yi cuts (3) in the right line 

A{x — xi) + C(z — 2i) = 0, or z — Zi = — -^ (x — x x ). 

Now if (3) is to be the tangent plane at P, this last equation 
must be that of the tangent line PR, and on comparing it with (1) 

we have — -pz — -r— . In a similar manner it can be shown that 
C dxi 

— -^ = -t— ^. Substituting in (3), we have as the equation of the 
C dyi 

tangent plane to the surface z = f(x, y) at the point (xi, yi, Zi), 

(t) g - Zl={x - Xl) ^ + ( y- yi) ^. 

The line normal (perpendicular) to the tangent plane at the 
point (xi, 2/1, 2i) is termed the normal to the surface at this point. 
It is the line NPN of the figure. By the principle of Art. 190, 

the direction cosines of this normal are proportional to ~, -^-, 

dxi dyi 

— 1, and therefore, by (VI), Art. 190, the parametric equations of 
the normal are 

dZi 
dxi 

dZi 



(n) 



x = xi + — r, 



y = y ' + e yi 

z = Z\ —r. 



The notation p = — , q = — , is frequently employed, and with 
ox cfy 

this notation the equations of the tangent plane and normal line 

have the forms 



298 DIFFERENTIAL CALCULUS §199 

(ti) z -z x = p t (x - Xi) + qx(y - y x ), 

(ni) x = xi + pir, y = yi + qir, z = z x — r. 

The actual values of the direction cosines of the normal are 
■ Pi gi -1 

Vi+pS + q/ Vi + pS + q/ VTTp7~+q?' 

Examples. 

1 . To find the equations of the tangent plane and of the normal to the 
surface 

xy + yz + zx = axyz 
at the point (xi, y i} Z\). 

Solution. The given equation can be thrown into the form 

1,1,1 

— I V- = a. 

x y z 

dZ Z 2 dZ Z 2 

Then — = -, — = , and the equation of the tangent plane is 

dx x 2 ' dy y 2 M 

*-*i = -ri(*-a;i)-^i(y-yi), 

^i yx 

which reduces to 

£. + X + JL = a . 
Xi 2 yx 2 Zi 2 

The equations of the normal are then 

x = xi + —r, y = yi + — r, z = Zt+ —r. 
x-c y\ Z\ l 

2. To find the equation of the tangent plane to the surface 

z = x sin y + y sin x 
at the point (x h yi, Zi), and of the tangent plane and normal at the point 



(!4-)- 



a 7 i- dZ . . dZ . . 

Solution. — = sm y + y cos £, — — x cos w + sin x. 
dx dy 

Hence the equation of the tangent plane at the point (x h y u s x ) is 

z — Zi = (sin y x + y x cos x x ) (x — Xi) + (xi cos ?/i + sin »i) (?/ — j/i). 

When Xi = yx = J , and Zi = tt, this equation becomes 

z - -k = x - ! + # - 1 or z + ?/ - z = 0. 



§§200-201 PARTIAL AND TOTAL DERIVATIVES 299 

Then the equations of the normal at this point are 

7T , 7T , 

z = 2 + r ' y ' = 2 ' = 7r-r - 

200. Exercises. Find the equations of the tangent planes and 
of the normals to the following surfaces at the point (x h y h z x ) : 

1. x 2 + y 2 + 2 2 = « 2 . 6. z?/ + yz + z:c = 0. 

2. -*±f-'±^=l. 7. ^ = 2 ^. 

a 2 o- c 2 

3. 2 = hxy. 8- 20 = e 1 . 

4. x 3 + ?/ + z 3 = a 3 . 9. jc! _|_ y l + 2 f = a f. 

5. x 3 — y 3 = a 3 . 

10. Find the equations of the tangent plane and of the normal to the 
surface z = sin xy at the points where x± = and j/i is any finite number, 

and also where Xi = -, tt, and y x is an integer. 

11. Find the equations of the tangent planes and of the normals to 
the surface z = sin (x + y) at the points (0, 0, 0), 1 |, ± |, J, y~, ^, 1), 

(TT, ±7T, 0). 



(t-I'°) 



201. Partial Derivatives of Higher Orders. If 2 is a function 

of re and y, — and — are, in general, themselves functions of x and 
dx dy 

y and may be differentiated partially, yielding partial derivatives 

of the second order. Thus — — is the second partial derivative 

dx dx 

d 2 Z 2 . d dz 

of z as to x, and is denoted by -£-* or D x z or z xx . Again, — — 

OX" (jy ()y 

d 2 Z 
is the second partial derivative of z as to y, and is denoted by — ^ 

°y 

D y z, or z yy . Further, — - — , denoted by- — —, and-;— — , denoted 
dy dx dy dx dx dy 

d 2 z 
by , are mixed second partial derivatives of z as to x and y. 



300 DIFFERENTIAL CALCULUS §201 

Now it can be proved, though the proof will not be given in this 
book, that 

d 2 z d 2 z 

dy dx ~ dx dy 

that is, that we get the same result whether we differentiate first 
as to x and then as to y, or first as to y and then as to x. Hence 
in the case of a function of two arguments there are but three 
second partial derivatives, 

d 2 z d 2 z d 2 z d 2 z 

dx 2 dy 2 dx dy ~ dy dx 

The mixed derivative is also symbolized by D 2 xy z and z xy . 

In case z is a function of three arguments, x, y, t, there are six 
partial derivatives of second order, viz., those given above, and in 
addition to them the three following: 

d 2 z d 2 z _ d 2 z d 2 z d 2 z 

dt 2 dxdt ~ dtdx dydt ~ dtdy 

The second partial derivatives may themselves be differentiated 
partially, yielding partial derivatives of the third order. Thus if 
z be a function of x and y, there are four third partial derivatives : 

d 3 Z d 3 z d 3 z d 3 z d 3 Z d 3 Z d 3 z d 3 Z 



dx 3 dy 3 dydx 2 dx 2 dy dxdydx dxdy 2 dy 2 dx dydxdy 

These identities are again assertions of the general principle that 
in taking a mixed partial derivative of any order, provided the func- 
tion be continuous, it is immaterial in what order the successive 
differentiations are performed. 

This principle we state without proof. 

If z be a function of the n variables x h x 2 , . . . ■ x n , the type of 
the partial derivative of the mth order is 

d^z 

dx^ dx^- dx^, . . . dx n r » 

where the r's are any positive integers which satisfy the condition 

ri + r 2 + . . . + r n = m. 



r 



§§202-203 PARTIAL AND TOTAL DERIVATIVES 301 

Example. When z = - , find the partial derivatives of z of the first 

y 



three orders. 




dz _ 1 dz _ 
dx y dy 


X 

y 1 



dx 2 dx\y) ' dy 2 dy[ y 2 ) y* ' 

dxdy dx\ y 2 ) y 2 ' r dy dx dy\y) y* 

3x 3 dx ~ ' a?/ 3 dy\y z ) y* ' 

„ =—0=0, or = — o=0; 

dydx 2 dy dxdydx dx\ y 2 / 

dh = d /2 x\ _ 2 or d 3 z d I 1 \ 2 

ax a?/ 2 dx\y z ) y 3 ' dy dx dy dy \ y 2 / ~ y z 



202. Exercises. 

1. Verify the principle of the mixed partial derivative for the third 
order in the following: 

(a) z = x 3 y — x 2 + 4 y z . (b) z = x ~~ y . (c) 2 = sin (x 2 y). 

x + y 

d 2 U d 2 U 

2. If u = x sin y + 2/ sin x, show that — - + — — + w = 0. 

dx 2 dy 2 

3. Find all the third partial derivatives of u when 

u = x sin (y + z) + y sin (x + z). 

x 

4. Find the second derivatives of y when y = tan -1 -• 

z 

5. Find all the second derivatives of z when x 2 + y 2 + z 2 = r 2 . 

6. Find all the second derivatives of z when log (y + z) — e* w = 1. 

203. The Total Derivative. Let 2 = f(x,y) and let z and y 
take simultaneously the increments Ax and A?/. Let Az be the 
resulting increment of z. Then 

(1) Az = fix + Ax, y + Ay) - /(x, y>." 



302 DIFFERENTIAL CALCULUS §203 

By subtracting and adding f(x, y + Ay) we throw this expression 
into the form 

(2) Ac = /(s+As, y+Ay) -f(x, y+Ay) Ax 

K Ax 

f(x,y+Ay)-f(x,y) . 

Ay y ' 

A little reflection will make it clear that the coefficient of Ax in (2) 

is equal to _ — + e,, and that the coefficient of Ay is 

ox L 

df(x y) 
equal to + yj, where e 1 and tq are infinitesimals. More- 

ox 

over, we assume that f(x, y) and its first partial derivatives are 

continuous, and therefore, from our definition of continuity in 

Art. 196, it follows that 

]im aJx 1 y+M ^J(ay}, whence ef(x,y + Ay) = d J(^y) 

Ay=o ox dx dx dx 2 

df(x v} 
Hence the coefficient of Ax in (2) is equal to + e x + e 2 . 

Setting e 1 + € 2 = e, equation (2) takes the form 

(a) Az = ^Ax+~Ay + eAx+riAy. 

Because x and y take increments simultaneously, this Az differs 
from the partial increments, A x z, A y z, that arise when only one of 
the variables, x, y, takes an increment. Az is termed the total 
increment of z. The variables x and y may be entirely independent 
or they may be functions of one or more other variables. 

Let us suppose for the present that x and y are functions of the 
single variable t. Then Ax and Ay depend upon At, and by divid- 
ing (a) by A£ and taking limits, we have 

T dz _ dz dx dz dy 

dt dx dt dy dt 

dz . 

37 is termed the total derivative of z as to t. It is the result of the 

at 

total change, Az, produced in z by the variation of t, and is made 



§203 PARTIAL AND TOTAL DERIVATIVES 303 

up of two parts, — -^> which is the result of the change produced 

uX (XL 

in z by the variation of t in x, and -~ -^> which is the result of 

the change produced in z by the variation of t in y. It is custom- 
ary to use the straight d for total derivatives, and the curved d 
for partial derivatives. The notation D is used for both total 

d t du 

and partial derivatives. The derivatives -^ and -r are total be- 
cause x and y are in this case assumed to be functions of the 
single variable, t. 

There is nothing new in formula I, for it is merely the formal 
statement of the successive steps of a differentiation, — steps with 
which the student is already familiar. The following examples 
illustrate this fact. 

Examples. 

1 . To find — when z = - and x and y are functions of t. 
at v 



§£ — I §£ - _ £ 
dx y' dy y 2 



We have — = -, — = — — , and on substituting in I we get 

dx dy 
v — — x ~ 
dz _ ldx _ x_dy _ dt dt 

dt y dt y 2 dt y 2 

which is the familiar rule for the differentiation of a fraction. 

dz 

2. To find— -when z = x z + x 2 y — y*, and x and y are functions of t. 

Differentiating this in the usual way, we have 

Now let us determine — by formula I. We have 
at 

f? = 3^ + 2^, f = x 2 -±y\ 
dx dy 

and on substituting in I we get the same result as before, which shows 
again that I is the formal statement of the steps of a differentiation. 

3. To find ^ when z = log (x 2 + 2 y), x = sin" 1 1, y = Vl -t\ 

at 



304 DIFFERENTIAL CALCULUS §203 



We have 




dz 2x dz 2 


dx 1 dy —t 


dx x 2 + 2y' dy x 2 + 2y' 


dt Vl-i 2 ' dt Vl-t 2 


Substituting in (I), we get 




dz 2x 1 2 


t 2(x-t) 



dt x 2 -\-2y Vl-£ 2 x 2 + 2y Vl-t 2 {x 2 + 2y)Vl 

dz 
and from this last result x and y may be eliminated and — be expressed 

at 

in terms of t only, if it is desired. 

In actual practice it is seldom necessary to separate the steps 
of the differentiation as was done in these illustrative examples. 
The differentiation should be performed in the usual way. Thus, 
in example 3 above 

dt x 2 + 2y x 2 + 2y 

2{x-t) p 

(x 2 + 2y) Vl -t 2 

dz 
When z is a function of x but not of y, — = 0, and formula I 

dy 

becomes 

t* dz dzdx 

1 Tt=Txdt or D*-D*D* 

which is the equivalent of formula (A) of Art. 84. Observe that 

when z is a function of x alone, — and -=- mean the same thing. 

If in I we set t = x, the formula becomes 

T dz _ dz dz dy 

a dx dx dydx 

If further in I a , z=f(x, y) = 0, then-r-= -r- = 0, and I a becomes 

j ^.^^ = or ^" + ^^ = 0. 

dx dydx ' dx dydx 



§203 PARTIAL AND TOTAL DERIVATIVES 305 

From this we get 



V 



dy _ dx 
dx df 

dy 

which is the equivalent of formula (B) of Art. 84, and is the formal 

dy 
statement of the process of finding -p when y is given as an implicit 

function of x by means of the equation f(x, y) = 0. 

Example. 

If x z -\- 3 x 2 y + y 2 = 0, we have, on differentiating in the usual way, 

3^ + 6, y + (3^ 2y) g=0, W hence| = -|fi|| = -|. 

dy 
The equations of the tangent and normal to the curve 
K x > y)= a ^ the point (x h wj we know to be 

y ~ Vl = dx 1 ( X ~ x ^ and y ~ yi= ~ ty x & ~ Xl ^' 

dx\ 

Substituting in these the values of -p= given by I'&, and reducing, 

we have as the equation of 

the tangent (y — wi) -^- + (x — Xi) -^ = 0, 
oyi oxi 

the normal (y — y{) — ^ (# — #i) ^- = 0. 

OX\ ot/i 

The foregoing formulae can be extended to functions of any 
number of variables. Thus if u be a function of x, y, and z, then, 
by a process of reasoning entirely similar to that by which (a) 
was obtained, it can be shown that Aw, the total increment of w, 
is given by the equation 

(a') Aw = -^ As + ~ At/ + -^ Az + eAx + tq Aw -f 0Az. 



306 DIFFERENTIAL CALCULUS §204 

And if further x, y, z are functions of the single variable, t, we have 
the total derivative of u given by the formula 

T , du _ dudx du dy du dz 

~di~dx1t' i "dydi' i 'Tzdi' 

And in general if u = f(x\, x 2 , . . . x n ) and if each a; is a function 
of t, then 

TT du _ du dx\ . du dx 2 . du dx n _ ^ du dxi 

~dt~ dx!~di + dx 2 ~dt + * * * + ~dx n ~dx ~ ff^dxi Ht' 

i=n 

or D t u = ^ D Xi u D t Xi. 

i=l 

204. Exercises. 

dz 

1. z = uv, u and v are functions of x; find — - • 

ax 

2. z — sin x cos y, and x = tan -1 1, y = tan -1 - ; find — • 

x 1 

3. ^ = tan~ J - , and x = log t, y = - j find D*z. 

2/ * 



4. 2 = we v + ve u , and it and vare functions of x; find — 



dz 
dx 

5. z = f(u, v), u and v are functions of x and ?/, and x and y are func- 
tions of t; find— • 

at 

du 

6. u = /(x, ?/, 2), and 2/ and z are functions of a:; find — • 

7. /(x, y, z) = 0, a(x, y, z) = 0, and y and 2 are functions of x; find 
^ and f! - 

8. If x 2 + 2/ 2 + z 2 + 2 x?/z = 0, and x + ?/ + z + 2 = 0, show that 

oj/ = _ (1 - y) (x - z) and dz = _ (1 - g) (y - x) < 
dx (1 — x) (2/ — 2) ' dx (1 — x) (2/ — 2) 

d?/ dg 

9. If /(x, ?/, z) = 0, z = tf>(x, ?/), and y is a function of x, find -p and — • 

10. If /(x, 2/, z) = 0, x = <f>(y, z), and 2/ and z are functions of x, find 

*and ^. " 

ax ax 

11. If x?/2 = 1, xy + 1/2 + 1 = 0, find — and — • 



§205 PARTIAL AND TOTAL DERIVATIVES 307 

205. Partial Derivatives of f(x,y,z, . . .), where x,y,z, . . . are 
themselves Functions of Several Independent Arguments. In 

the preceding article all the functions involved depend ultimately 
on a single variable, as t. Let us now suppose that z = f(x, y) and 
that x and y are functions of two independent arguments r and s; 
that is, suppose x = 4>(r, s), y = x(r, s). If we hold s fast and 
let r take an increment Ar, x, y, and z will take partial increments 
A r x, A r y, A r z, and, employing the same argument used at the 
beginning of Art. 203, we shall arrive at an equation which differs 
from (a) only in having partial increments where (a) has total 
increments. And by holding r fast and allowing s to vary, we get 
a similar equation. These equations are 



(b) 



A r z = — A r x + — A r y + eiA r £ + rnA r y t 
A 8 z = —A s x + y A 8 y + e 2 A 8 x + f] 2 A 8 y. 



Dividing these equations respectively by Ar and As and taking 
limits, we have 

dr dx dr "*" By dr ' ds ~ dx ds dy ds ' 

formula? which differ from I only in having all their derivatives 
partial. Using the symbol D for derivatives, we may write 

(A) . D t z = D x zD t x + D v z D t y, 

which includes I and III because D a may mean either a partial or 
a total derivative. 

If y does not occur in z, then-— = 0, and III becomes 

dy 

-rj-r* dz _ dz dx , dz _ dz dx 

dr~ dx&r and ds~te as"' 

Compare I* of Art. 203 and formula (A) of Art. 84. 
Again, in III let x = r; that is, suppose z = f(r, y) and 



308 DIFFERENTIAL CALCULUS §205 

,, v ,, dx dr 1 dx dr „ 

y = 0(r, s) : then -r- = v = l>^ = ^ =0 (because r and s are 
* K ' dr dr ds ds 

independent), and III becomes 

TTT dz _ fdz\ dz dy , dz _ dz dy 

dr \drl dy dr ds dy ds 

Here we have two kinds of partial derivatives. The one in brackets 

— J is due to the variation of r in z outside of y as it were, the 

other, — , is due to the total variation of r in z, both outside and 
inside of y. 

r 
Example. Let z = - and y = s sin r. 

y 

m . fdz\ 1 dz r dy dy , ... ,. 

Then [ — = -> — = : : -^ = s cos r, -^ = smr. Substituting in 

\dr y dy y 2 ' dr ' ds 6 



(i) 



Ilia, we have 



dz 1,1 r\ y — rs cos r sin r — r cos r 

— = -+ -scosr = = — ; 

dr y \ y 2 / y 2 s sm 2 r 



y + \ yV 



dz r \ . r smr r 

— smr= — = — ——. 

ds \ y 2 / y 2 s 2 smr 



In this example we have separated the processes of finding the partial 
derivatives into their several steps, in order to illustrate the formula. 
In actual practice, however, the differentiation would be performed thus : 

y — r — 
dz dr _ y — rs cos r _ sin r — r cos r 

dr 



yl 




y 2 


s sin 2 r 


dz 


Jidy = 
y 2 ds 


r sinr 

y 2 


r 


ds 


s 2 sinr' 



This example may be solved more directly by first substituting in z 
the value of y given by the second equation. Thus 

r dz sin r — r cos r dz r 

z = 



s sin r dr s sin 2 r ds s 2 sin r 

And this method is preferable whenever the substitution can be easily 
made. 



205 



PARTIAL AND TOTAL DERIVATIVES 



309 



Formula III can be extended. Thus, let u = f(x, y, z) and let 
x, y, z be functions of r and s. Then it can be shown that 



(bo 



. du A . du . . du A . . . A . . . 

Ku = — A r x + — A r y + -7j- A r z + eiA r z + rnA r y + 0iA r z, 

A du A . du A . du > . A . A . " A 

A s w = — A fi z + — A 8 y + — A s z + e 2 A a z + Y) 2 A a ?/ + 2 A 8 z. 



From these we get 
III' 



du _dudx_ dudy du dz . dw _ du dx dudy du dz 
dr dx dr dy dr dz dr ' ds dx ds dy ds dz ds 



And in general let 

z = Kvi, 2/2, .. . y n ) 

and yi = 0»(ri,r 2 . . . r k ), where i = 1, 2, 



. n. 



Then by an argument entirely analogous to that by which the 
preceding formulae have been established, it can be shown that 

f^L = ^i_ f}Ml _j_^_ ^?i. d^dyn 

dr x dyi dr x dy 2 dr x ' d?/ n dn ' 



IV 



dz = dz gg/i dz a^ 2 

dr 2 d?/i dr 2 dy 2 dr 2 



, dz_ dyn 
dy n dr 2 



dz_^dz_ dyi dz dy 2 

dr k ~~ dyi dr k dy 2 dr k 



+ 



dz dy n 
dy n dr k 



which may all be united in the single formula 



(B) 



£-££*£ -^-S*** 



(s = 1,2,3, k). 



The last formula of (B) includes all the formulae I, IV. 

Formulae III, IV, (B), like formulae I, II, (A), 

are merely the formal statements of the successive steps of a 
differentiation. 



310 DIFFERENTIAL CALCULUS §206 

206. Exercises. 

When possible solve by each of the three methods explained in the 
preceding example. 

1. z — x 2 + y 2 , x = r — s + 3, y = r + s — 3. 

Show that — = 4 r, and — = 4 (s — 3) . 
dr ' ds 

2. z = tan- 1 - , x = s sin t, y = V s 2 + t 2 . 

V 



Show that s — + t — = 



dZ dZ _ yst COS t 

ds + t dt~ x 2 + y 2 

3. u = xy + yz -\- zx, x = r cos d, y = r sin 0,2 = r0. 

Show that — = — ■ • 
dr r 

4. If u is a function of z, y, and 2, and if 

x = air + &is + Ci£, ?/ = «2^ + b 2 s + c 2 £, 2 = a z r + b 3 s + c 3 t, 

n . dU dU dU 

5. 2 = sin -1 - , w = log rs: find — and — . 

6. w = /(z, y),x = r cos d, y = r sin 0. 

7. u = tan -1 - , z = r cos 0, w = r sin 0: find — and — in terms of r 

x 1 ' dr dd 



and 0. 



If u = f( U \ show that x ~ + ?/ ? = 



ax 
#in£. Set 2 = - > and apply formula III*. 

9. If w = fix + ax', w + by'), show that — , = a — » and — : = 6 — - 
JK ' u y " dx' dx dy' dy 

10. If u = f(x + at, y + &0j show that — = a \- b 

dt dX dy 

11. If u = xyfl- ), show that x \- y — = 2 u. 

\xj dx dy 

12. If f(ax + by) = c. find^- 

aa; 



§207 PARTIAL AND TOTAL DERIVATIVES 311 

13. If {ax + by) 2 + tan (ax + by) = c, find -f-. Compare with exer- 

ax 

cise 12. 

14. If z = 4> (x 2 - y 2 ), prove that y — + x — = 0. 

15. If z = 1// K V prove that z-+2/-+z = 0. 

x \xj dx dy 

207. Implicit Functions. We shall consider some special cases 
of the foregoing formula?. 

1 . Let u = f(x, y, z) and let sbea function of x and y ; to find 

du , dw 
— and — • 
5a; dy 

The desired results are gotten by setting r = x and s = j/ in 

III' of Art. 205, noting that then |? = |^ =1, ^ = |^ = 1, 

dr 6a; ds dy 

and that, because z and ?/ are independent, -^ = -^ ( = 0, and 

|2 = |5 = o. We get 

TTT , dw _ /du\ du dz , dw _ /diA dw dz 

lila Tx ~ \~dx~) ^ Tz di ancl ^~U^ + d?V 

where the bracketed derivatives have the meaning explained in 
Art. 205. 

Query. What is the difference between III a and III '? 
Problem. Obtain III ' from equation (b') of Art. 205. 

The following formula III& for the differentiation of implicit 
functions is a special case of Ilia'. 

2. Let u = f(x, y, z) = 0, and let x and y be independent as 

in 1 above. Then — = 0, — = 0, and Ilia' becomes 
dx dy 

Of , df dz _ , df , df dz A 

III 6 3T + "/ IT = ° and T^ + / T- = Of 

dx dz da; dy dz dy 

which are analogous to I& of Art. 203. 



312 DIFFERENTIAL CALCULUS §207 



From III& we get 

M 

ttt / dz dx , dz 
dx df dy 
dz 


df 

dy 
df 

dz 


which are analogous to V of Art. 203. 





In Illb and IIV the brackets are omitted from about ~ and 

dx 

df 

■/- because no misunderstanding can arise from this omission. 
dy 

3. Let f(x, y, u, v) = 0, and g(x, y, u, v) =0 be two equations 
in the four variables x, y, u, v. These equations serve to deter- 
mine two of the variables, which we take for functions, in terms 
of the other two, which are therefore arguments, and which are 
independent. We choose u and v to be functions, and seek expres- 
sions f or — - i — > — i — . The partial derivatives of / and q as to 
dx dy dx dy J y 

x and y are 0, and therefore by formula (IV) or (B) of Art. 205 

we have 

df 4 _dl§udfdi = dg ,dgdu c^cb 

dx "*" dudx^ dv dx ' dx ^ dudx^ dv dx ' 

d l\^L^jL.^l^l = o and ^ + -^ — + ^—=0 
dy dudy dvdy ' dy du dy dvdy ' 

These equations may be solved, the first two for 

— and — , and the last two for — and — • 
dx dx dy dy 

In solving such problems as this, as well as in the application 
of any of the preceding formulae, the essential thing is to have 
fully in mind at every step just what variables are functions and 
what are arguments, and, in taking derivatives as to any of the 
latter, to regard all the other arguments as constants. Some- 
times there is a liberty of choice as to which variables shall be 
regarded as functions, and which as arguments. In the above 
example, for instance, we could have considered x and u to be 
functions, and y and v to be arguments, and other choices could 
have been made. 



§ 208 PARTIAL AND TOTAL DERIVATIVES 313 

Example. Given u 2 — v 2 + x 2 — y 2 = 0, uv + xy + 2 = 0, to find — » 

dV dU dV 

dx dy dy 
Differentiating each equation as to a:, we have 

dU dV . A , dU . dV . n 

w y [- x = and y — + w — +2/ = 0. 

a# az az dx 

Solving these for — - and — j we have 
dx dx 

du _ _ ux -j- vy , 3y _ vx — uy 
dx u 2 + v 2 dx u 2 -f- y 2 

Differentiating the given equations as to 2/, we have 

du dv A i du . dv . n 

w • y — — y = and y- f-M f- £ = 0. 

Solving for — and — j we have 
dy dy 

du _ uy — vx i dv _ _ ux -f vy 
dy u 2 + y 2 a?/ w 2 + y 2 

208. Exercises 

1. In the last example find the four partial derivatives when the two 
functions are 

(a) u,x. (b) u,y. (c) v,x. (d) v,y. (e) x,y. 

2. Differentiate w = /(#, 2/, w, v) under each of the following hypoth- 
eses : 

(a) x, y independent; u, v functions of both x and y. 

(b) x, y independent; u a function of x and y } v of y only. 

(c) x, y, u independent; v a function of x, y, and u. 

(d) x, y, u independent; v a function of x and u only. 

(e) a; independent; y, u, v functions of x. 

3. w = /(w, y), w and y are functions of both x and y, and z and ?/ are 

functions of t. Write a formula for— — which shall contain: 

dt 

(a) derivatives as to u, v, t only; 

(b) derivatives as to x, y, t only; 

(c) derivatives as to u, v, x, y, t. 

4. ux -\- vy -\- 1 = and u-\-v + x + y = 0. Show that 

^i^_ = 0-— 4- — =o- d 2 ^ , a 2 y _ 
ax 2 az 2 ' dy 2 dy 2 ~ ' ax ay dx dy ~ 



314 DIFFERENTIAL CALCULUS §209 

5. z 2 - x 2 + y = 0, x 2 - r 2 + s = 0, y 2 - s 2 + r = 0. Find deriva- 
tives of z. 

6. Vz + Vx + Vy = !j Vz + Vr ~ VI = h Vy ~ Vr — Vs = 0. 
Find the derivatives of z. 

7. Differentiate u 2 + 2 vx + 2 ?/ = 0, y 2 + 2 w/ + 2 x = 0, 

(a) when x and ?/ are independent; 

(b) when u and y are independent; 

(c) when x and u are independent. 

8. x=f(u,v),y = g(u,v). Rnd _,_,_,_. 

9. w 2 + y 2 — £ 2 = 2 ?/, m + v — y = x. 

Find the first and second derivatives of u and v as to x and ?/. 

10. If f(x, y) = 0, 4>(x, z) = 0, show that ^^ = ?L?±. 

JK ' J ' v ' ' dxdydz dx dz 

11. If/(p,»,0=0, show that ^ — —=-1. 

12. If f(x + y + z, x 2 + ?/ 2 + 2 2 ) = 0, show that 

fa - s) rr + (2 - x) t- = a; - y. 

ox dy 

13. If /( z 2 — xy, - ) = 0, show that xz — - + yz — = xy. 

\ x) dx dy 

14. If z = f{x + y) + g(x - y), show that ■— - = — ■ ■ 

dx 1 dy 1 

15. If e 2 = xf(-\ show that 



i. If e 2 = xf(-\ show 

\y) 



dz . dz ., i dx , dx 

x — + y — • = 1 and y = x. 

dx dy dy dz 

209. The Tangent Plane and the Normal Line. By means of 
IIV of Art. 207 the equations of the tangent plane and normal 
line can be written in forms that are more symmetrical and con- 
venient than were those given in Art. 199, where z was given as 
an explicit function of x and y. We there obtained the equation 
of the tangent plane in the form 

( s dZi t , . dzi 

If the equation of the surface be written in the form f(x, y, z) = 0, 



§210 PARTIAL AND TOTAL DERIVATIVES 315 

dz dz 

— and — are given by formula IIV, and on substituting their 

values there found, in the equation of the tangent plane just given, 
and reducing, the latter takes the symmetrical form 

(t) (x - Xl) £ i + {y - yi) ^ + {z - Zl) M = . 

The direction cosines of the normal to the surface f(x, y,z) = 
at the point (x h y h Zi) are therefore 

#-+VS, f + VR, f + VR, 

dxi dyi dzi 

l 

Hence the parametric equations of the normal are 

(n) X = Xl +Jx~ 1 r > y = Vl + ^ Z = Zl + ~dz~J' 

Example. Let us find the equations of tangent and normal to the 

ellipsoid — + f- + -„ = 1 at the point (x h y h z x ). 
a 1 b l c z 

x 2 u 2 z 2 j 
Solution. In this case f(x, y,z) = — + f- + - - — 1 = 0. 

cr o z c l 

Hence 

df = 2x d£ = 2y d£ = 2_z 

dx a 2 ' dy b 2 ' dz c 2 ' 
and the equation of the tangent plane is 

(x-x l )^+(y-y l )^ + (.z-z l )f 2 = 0, 

which reduces to 

ft 2 T b 2 I" C 2 L ' 

The equations of the normal may be written down at once, 

i X\ , Vi . Z\ 

x = xi + - 2 r, y = yi + y -r, z = z 1 + -r. 

210. Differentials, Partial and Total. Let z be a function of two 
independent variables, x and y. We define the partial differential 



316 DIFFERENTIAL CALCULUS §210 

of zastox to be the product of the partial derivative of zas to x by the 

dz 
increment of x, and shall denote it by d x z, so that d x z = — As. 

dx 
When z = x, d x x (= dx) = — Ax = Ax. 

ox 

We may therefore replace A a; by dx in the above definition, and 
have as our definition the equation 

(1) d x z = — dx; and similarly, d y z = ~^~dy. 

ox oy 

From this we get by division 

dz_ _d x z , dz __ d y z 
dx dx dy dy ' 

which constitute a new notation for partial derivatives. When z 
is a function of x alone, or of y alone, definition (1) is virtually the 
same as that given in Art. 73. 

We shall next define the total differential of z to be the sum of 
its partial differentials. If we denote it by dz, we have 

dz = d x z + d y z, 
or 

(C) <fe = gete + g<%. 

It is to be kept in mind that we have thus far assumed x and y to 
be independent. Were they not so we could not write Ax = dx 
and Ay = dy. When z is a function of a single variable, or when 
z is itself independent, its only differential is of course a total 
differential. 

Suppose now that z is a function of x and y, and that x and y 
are functions of the independent variables r and s. Then z is of 
course a function of r and s and by (1) and (C) we have 

(10 d r z = — dr, d s z = — ds, dz = —dr + ■— ds. 

v dr ds dr ds 

But in this case z has two other partial differentials, viz., the par- 
tial differentials of z as to J , which we define to be the product of 



§210 PARTIAL AND TOTAL DERIVATIVES 317 

the partial derivatives of z as to ! by the total differential (not incre- 
ment) of j j , and denote by d x z and d v z. We have then 

(2) d x z = — dx and d y z = — cfy. 

dx dy 

This definition includes (1) above. It holds, too, if x and y are 
functions of one variable only. We shall now show that (C) holds 
also when x and y are no longer independent. 
By III of Art. 205, we have 

TTT dz dz dx , dz dy . dz dz dx , dz dy 

III — = ^-^- + ^-^ and — = ^--t-H-^-^' 

dr dx dr dy dr ds do: as dy ds 

Substituting these in the last equation of (1') and rearranging 
terms, we have 



dz 



dx\dr ds ) dy\dr ds ) 



But by definition (C), the quantities within the brackets are dx and 
dy, and consequently (C) holds when x and y are functions of r and 
s as well as when they are independent. It holds of course when 
x and y are functions of one variable only. 

Let us now substitute in the first two equations of (1') the 

values of — and — given by III. The result is 
dr ds 

1 dz dx , , dz dy 7 

d r z = — — dr + — -f- dr, 
dx dr dy dr 

and a similar equation for d s z. But by (1) 

— dr = d r x, -f-dr = d r y, 
dr dr 

and similar equations for d s x and d 8 y. 
Hence 

(3) d r z =-^-d r x + — d r y and d s z = — d s x + -5- d s y. 

dx dy dx dy 

These formulae differ from (C) only in that all the differentials of 
(3) are partial. We may therefore omit to distinguish by sub- 



318 DIFFERENTIAL CALCULUS §210 

scripts partial differentials from total, and may regard (C) as 
covering every case, — the case in which x and y are independent, 
that in which they are functions of one or of several variables, 
that in which the differentials are all total, and that in which 
they are all partial. 

The matter will be made clearer by considering the case where 
z = f(u, v), u and v are functions of x and y, and x and y are 
functions of the independent variables r and s. Then 

(°'> *-£** + S* 

where a and jS may be u and i;, or a; and y, or r and s, and the 

differentials involved (dz included) may be all total or all partial. 

Problem. Write down the nine forms which (C) may take in this case. 

Formula (C) is an important one because it- is equivalent to 
formulae I . . . Ill of the preceding articles, and because those 
f ormulae can be derived from (C) . Thus, dividing (C) by dt gives 
I of Art. 203; dividing by dx gives I a or I 6 ; or, regarding x 
and y as functions of r and s and dividing first by dr and then 
by ds, and writing d instead of d, we get III of Art. 205. 

Formula (C) is somewhat more comprehensive than are formulae 
I . . . Ill of the preceding article; for (C) holds by definition 
when x and y are independent variables, whereas no derivative 
equation can be obtained from (C) in such a case. However, when 
x and y are independent, (C) is of very little use; for a differential 
formula is serviceable only when derivative formulae can be de- 
duced from it. 

Formula (C) can be extended to functions of any number of 
arguments. Thus, if z =f(yi,y2, . . . y n ), then 

( D ) dz = JLfr+JLdg, + ■■■ + ^ d y n J^d yi , 

where the y's may be independent or dependent upon other 
variables, or some of the y's may be independent and the others 
dependent, and the differentials (dz included) may be total or 
partial. 



§§211-212 



PARTIAL AND TOTAL DERIVATIVES 



319 



Exercises. 

1. z = f(x, y) and y is a function of x. Write the formulae for d x z 
and dz. 

2. z = f(r, s, y) and y is a function of r and s. Write the formulas 
for djZ, d 8 z, d y z, and dz. 

3. z = f(r, s, x, y) and x and y are functions of r and s. Write the 
formulae for d&, d 3 z, d x z, d y z, and dz. 

It will be observed that finding the above differentials is exactly like 

finding the corresponding partial derivatives — , — , — ■ , etc. In operat- 
ing with differentials there is seldom any need to distinguish by subscripts 
partial from total differentials. 

211. Remarks. We repeat here what was said in Art. 75, 
that small advantage accrues from the use of differentials. They 
are relics of the early days of the Calculus and, as the reader is 
probably now ready to admit, constitute for the beginner an 
obstacle to the understanding of the Calculus rather than an aid. 
Nevertheless they are in almost universal use and it is therefore 
necessary that the student of mathematics become thoroughly 
familiar with them. 



212. Geometric Representation of Partial and Total Differ- 
entials. Let f(x, y) be a function of two independent variables, 
and suppose the surface whose equation is z = f(x, y) to be 
drawn. Let P be the point (x, 
y, z) and Q the point (x + Ax, 
y + Ay, z + Az) upon this sur- 
face. As in the figure of Art. 
199, the partial increments 
A x z, A y z are AP h BP 2 . It is 
plain from the figure that the 
total increment Az is CQ. Ob- 
serve that the total increment 
is not in general the sum of the 




partial increments. 



We saw in Art. 199 that — and — were the 

dx By 



320 DIFFERENTIAL CALCULUS §212 

slopes of the tangent lines PR and PS. Therefore from our defini- 
tion of differential and from the figure it follows that 

d x z = — d# = AR and d v z — -r-dy — BS. 
dx dy 

That is, the partial differentials are the increments of the z-coordi- 
nates of the tangent lines PR and PS. 

PSP'R is the tangent plane to the surface at P, and a study of 
the figure will make it evident that 

CD =AR = d x z, and DP' = BS = d v z, 
and that consequently 

dz = d x z + d y z = CP'. 

In words, the total differential is the increment of the z-coordinate of 
the tangent plane. 

Compare this with the geometric representation of the differ- 
ential of a function of a single variable given in Art. 73. 



CHAPTER XXIX 



MULTIPLE INTEGRALS 



213. The Volume under a Surface. Consider the problem of 
finding the volume of a truncated right cylinder included between 
the #2/-plane and the surface 

•z =f(*,y). 

Let the base of the cylinder be R, 
whose perimeter C may be a single 
curve or may be made up of parts 
of several curves. We have already 
learned that the volume in question 
may be found by cutting the solid 
into slices by planes parallel to one 
of the coordinate planes and taking 
the sum of the volumes of the slices. 
The volume of a slice such as that 
shown in the figure may be denoted by A (x) dx, and then 




V = f h A(x)dx. 



To find A(x) requires an integration; for A(x) is the area of 
the plane section AiA 2 AiA 2 '. The equation of the curve AiA 2 
is z = f(x, y) , where x is constant. If we denote the ^/-coordinates 
of Ai and A 2 by 7 and 5, we have 



A(x) 



Jpy=s 
f(x, y) dy, 
y=y 



x remaining constant throughout the integration. 7 and 5 are 
functions of x determined by the equations of (C). Indicating this 

321 



322 



INTEGRAL CALCULUS 



§214 



by writing them y(x), 8(x) and substituting the foregoing value 
for A (x) in the formula for V, we have 

x =b y = S(x) x=b y=h(x\ 

V = I I /0, y) dydx = j J zdy dx. 

x = a y = y(x) x=o y=y(x) 

Here we have the volume given by two successive integrations. 
In the first integration, the inner one, x is to be considered con- 
stant, and after substituting the limits which are themselves func- 
tions of x, the resulting function A (x) is to be integrated as to x. 

By cutting the truncated cylinder into slices parallel to the xz- 
plane and proceeding as before, we get for V the formula 

y =d x=0(y) y = d x=0(y) 

V = / / f(x, y)dxdy = I j zdx dy. 



y=c x=a{y) 



y=c x=aiy) 



214. The Double Integral. Let fibea region of the a^-plane 
whose boundary, C, is a single closed curve, or is made up of parts 

of several curves. We suppose 
f(x, y) to be continuous at every 
point of R. Let R be cut up in any 
way into small curvilinear polygons. 
This can be done by drawing in the 
zz/-plane two or more sets of quasi- 
parallel curves as shown in the fig- 
ure. C itself forms part of the 
boundaries of those polygons that 
are adjacent to C. Denote the areas of the polygons by ARi, 
AR 2 , .... Let (x k , y k ) be any point in the interior of Ai4 and 
form the product f(x k , y k )^Rk. Form such a product for each of 
the small polygons, and take their sum 

f(x 1} y 1 )AR 1 +f(x 2 , 2/ 2 ) A£ 2 + . . . = Vf( X} y) A/2, 

R 

where R is written under the ]v to indicate that the summation is 
to extend over the entire region R. Now let the number of poly- 




§214 



MULTIPLE INTEGRALS 



323 



gons be indefinitely increased, but in such a way that the greatest 
diameter and consequently the area of each small polygon shall 
have the limit 0. Then the fundamental theorem of the Integral 
Calculus of Functions of Two Arguments is 

j£jf(pc 9 y)AR has a limit, and this limit is independent of the 

R 

method of subdividing R into small polygons. 

This theorem is an analytic one, but we shall give a geometric 
proof of it. 

On R as a base we construct a truncated right cylinder like that 
of the preceding article, whose upper base lies in the surface z = 
f(x, y). On each of the small poly- 
gons AR we construct small trun- 
cated right cylinders.* Plainly, the 
large cylinder on base R is the sum 
of the small cylinders. Consider one 
of these small truncated cylinders 
ABC DM, and the corresponding pro- 
duct f(x k ,y k )AR k . This product is 
the volume of a cylinder not trun- 
cated, A'B'C'D'M, whose base is AR k 
and altitude f(x k , y k ), and which co- 
incides with the truncated cylinder 
except in the small upper portion. 
Now it is fairly obvious that the limit 
of the sum of such cylinders as 
A'B'C'D'M is V, that is, that 

\im^f(x,y)AR = V. 

R 

As soon as this has been proved our theorem is proved. 

* We term such a small solid a cylinder, although its lateral surface con- 
sists in general of parts of several cylindrical surfaces, rather than of a single 
closed cylindrical surface. 




324 INTEGRAL CALCULUS §214 

The proof is as follows: 

Let v h v 2 , . . . be the volumes of the small truncated cylinders. 
Then 

V = 2) v and V = lim V v. 

R R 

Let z and Z be the smallest and greatest 2-coordinate in the trun- 
cated cylinder ABCDM and let v k be its volume. Then 

zAR k = v k ^ ZAR k . 

Let z k =f(x k ,y k ), and divide the inequality by z k AR k . There 

results 

z_ ^ v k ^ Z 
2a z k AR k z k 

Since by hypothesis /(#, ?/) is continuous, 2, ^, and Z all have the 
same limit, which is not 0; and therefore 

1 ^ lim— ^- ^ 1. 
z k AR k 

Hence lim — t-=- = 1, and by Duhamel's Theorem, Art. 173, in 
Zk&Kk 

the equation V = lim x ^, the y's may be replaced by the prod- 

R 

ucts Z&AZ4 or f(x k , y k )AR k , and therefore 

F = lim ]£/(*, &)A#. 

Moreover, it is plain that the above proof does not depend in the 
least upon the way in which R is subdivided. Hence our theorem 
is proved. 

The limit, lim X/fe 2/)Ai2, ls defined to be the double inte- 
r 
gral of f(x,y) over the region R, and is denoted by the symbol 

Jff(x,y)dR. 

r 
so that we have 

fff(*> y) dR s lim 2)A*« y)* B - 



§214 MULTIPLE INTEGRALS 325 

In the preceding article we found an expression for V, that is, for 
this limit, and may therefore write 

x=b y=5(x) y=d x=0(y) 

(A) j jf{x, y)dR=J jfix, y) dy dx = J Jf(x, y) dx dy. 

R x=a y=y{x) y=c x=a(y) 

It is customary to call any one of these three expressions the 
double integral of f(x, y) over the region R, but strictly speaking 
it is only the first one that is the double integral, because it is 
this one only that stands for the limit of the sum of products. 
The other two expressions are merely symbols that two succes- 
sive integrations are to be performed. In the double integral 

/ / f( x > y) dR, the integration signs and the differential dR are 

R 

employed not because limx/fc y)AR in itself suggests a process 

R 

of integration, for it does not, but because it is found that the 
value of this double integral can always be obtained by two suc- 
cessive integrations. 

We recapitulate our results in the following: 

Theorem. Letf(x, y) be continuous throughout a region R, and let 
a sum of products be formed as explained above. Then as the num- 
ber of products is indefinitely increased in any way, provided only that 
each product has the limit 0, the sum of products £.f(x, y)kR has 

a definite limit denoted by I I f{x, y) dR and termed the definite 

R 

double integral of f(x, y) over the region R. Moreover, the value of 
this double integral can be found by two successive integrations, the 
limits of these integrations depending upon C, the contour of R. 

Instead of dividing R into curvilinear polygons by curves, we 
may divide it into rectangles by straight lines parallel to OX and 
OY. Then, instead of small cylinders we shall have in the summa- 
tion small rectangular prisms, and if the distances apart of the 
parallel lines be denoted by Ay h Ay 2 , . . . Axi, Ax 2 , . . ., the 
volumes of these prisms will be of the form f(x, y)kykx, and we 
shall have 



326 



INTEGRAL CALCULUS 



§214 



x=b y=5(x) 



(B) lim^f(x, y)Aykx = / J f(x, y)dydx = / jf(x,y)dydx. 

R x=a % 

y=S(x) 

From this point of view the integral / f(x, y) dx 



y=y(x) 



where both x 



■y(x) 



and dx are constant, gives the sum of the volumes of the prisms 
that make up the slice A1A2A1A2 whose altitude is dx (see figure, 
Art. 213). 

Having seen that the double integral may be interpreted 
geometrically as the volume of a portion of a cylinder, it is now 

easy to see that any volume 
may be expressed as a double 
integral. For, let V be a solid 
bounded by a closed surface S, 
as shown in the figure. About 
the solid circumscribe a cyl- 
inder with axis parallel to OZ, 
and let R be the base of the 
cylinder in the xy plane. (The 
dotted curve is the curve of 
contact of the cylinder with V.) 
Then the volume of V is plainly 
the difference between that 
part of the cylinder which is 
cut off by Si, the upper part of the surface S, and that part which 
is cut off by S 2 , the lower part of S. If z = }i{x, y) and z = f 2 (x, y) 
be the equations of $1 and $2 respectively, we may write 

V = I ffi(x, y) dydx - J I f 2 (x, y) dy dx = J I (ft -/ 2 ) dy dx, 

R R R 

which was to be proved. 

Example. Let us find by double integration the volume of the sphere 

(x - a) 2 +(y- b) 2 + (z - c) 2 = r 2 . 
In this case R is the circle (x — a) 2 + (y — b) 2 = r 2 , and the limits of 
integration for y are 

y = b — Vr 2 —{x — a) 2 = y{x) and y = b + Vr 2 — (x — a) 2 = d(x). 




§215 



MULTIPLE INTEGRALS 



327 



Moreover, the equations of Si and S 2 are 



z = c + Vr 2 - (x - af - (y - 6) 2 = f x {x, y) , 



and 



z = c - Vr 2 - (x - a) 2 - (y - b ) 2 = f 2 (x, y ) . 
Hence ° 

x = a+r y=S(x) 

V x =j f[ c + ^ r *-( x - a y-( y -by]dydx, 



a a+r 



x—a— 


r y=y(x) 










x=a+r 


y=d(x 


and 


v 2 


-s 


Jl- 


Hence 




x=a—r 


y=y{x) 



dx. 



x=a-\-r y=S(x) 

V=V 1 -V 2 = ff(f l -f 2 )dydx = 2J f Vr 2 - (x - a) 2 -(y- 6) 2 dy dx. 

R x = a—r y=y{x) 

Integrating by the table of integrals, we have 

y=8(x) 



f Vr 2 - (x - aY - (y - b) 2 dy = | (y -b)Vr 2 -(x-a 



y-{y-by 



y=y{x) 



| 2/ =6+V / r 2 -(x-a) 2 



+ ir2 _ (a; _ a ) 2 ( s i n -l y ° = [r2 _ (a: _ a)2] . 

Vr 2 — (a: — a) 2 J y =6-Vr 2 -(x-a) 2 ^ 
Therefore 



x=a-\-r 



a+r 



V = 7T Mr 2 — (a; — a) 2 ] dx = v \r 2 x — - (x — a) 3 = -ttt 3 . Aws. 



215. A Plane Area Expressed as a Double Integral. In 

formula (B) let /(«, ?/) = 1. Then the double integral takes the 
z 




special form 



J> 



dx and is rep- 



resented geometrically by the vol- 
ume of a right cylinder of base R 
and of altitude 1. But because 
the volume is in this case expressed 
by the same number as the area of 
the base, we may regard this double 
integral as giving the area of R. 
From this point of view the area of 



328 



INTEGRAL CALCULUS 



216 



R is the limit of a sum of rectangles of dimensions Ay and Ax. 
We have then the general formula for a plane area, 

x=b y=8(x) 



R = f I dy dx = I j dy dx. 



Now 



and 



R 

S(x) 

I dy - 8(x) 
y(x) 



x=a y=y(x) 



y(x) = AiAi, 



nb nb 

R = / 8(x) dx — I y(x) dx = aAA 2 Bb — aAA x Bb. 

U a J a 

From this it appears that this process amounts after all to an 
application of our familiar formula for a plane area, A = I y dx. 

a 

In fact, there is seldom any advantage in using a double integral 
rather than a single integral to calculate a plane area. 

Problem. Express as a double integral in polar coordinates, p, 6, the 
area of a sector bounded by a curve and two radii vectores. 

Suggestion. Divide the sector into curvilinear quadrilaterals by radii 
and concentric circles. 

216. The Area of a Surface Expressed as a Double Integral. 

The area of a curved surface can be expressed as a double in- 
tegral extended over a plane region. To establish the formula 
we shall have need of the following 

Lemma. Let A be any area in a plane M, and A' its (orthog- 
onal) projection upon any other plane M f . 
Let 6 be the angle between the planes. 

Then 

A' = A cos 0. 

To prove this, we first cover A with a net- 
work of small rectangles by drawing lines 
parallel and perpendicular to L, the line 
of intersection of M and M'. This net- 
work projects into a similar network of rectangles covering A'. 
Let a be one of the rectangles of A, and a' its projection in A r . 




216 



MULTIPLE INTEGRALS 



329 



It is easily proved that a' = acos0, and from this it follows that 
lim Va' = cosfllim^a, or A' = Acosd. Q. E. D. 

Now let the equation of the curved surface be f(x, y, z) = 0, 
and let S be a portion of this surface bounded by a closed contour C. 
We seek a formula for the area of S. 

Let S be projected into a region S' of the xy-plsme. We assume 
that the perpendiculars upon the x?/-plane which project S into 
S' meet S in but one point; that is, we 
assume z to be a single-valued function of 
x and y throughout S'.* Circumscribe 
about S a polyhedral surface P, by con- 
structing tangent planes to S. As the 
number of tangent planes is indefinitely 
increased, the number of faces of P is 
indefinitely increased, and the area of P 
approaches as a limit the area of $.f Rep- 
resenting the faces of P by APi, AP 2 , • • • , 
it is fairly obvious that 




S = lim J) AP. 



The faces of P project into a network of polygons covering S'. 
Let (x, y, z) be the point of contact of the plane of AP with S, 
and AS' the projection of AP in S'. Let 6 be the angle between 
the plane of AP and the xy-plsme. Then by the lemma 

AS' 



AS' = AP cos 6, whence AP = 



cos d 



Now 6 is equal to the angle between OZ and the normal to S at 
(x, y, z) ; that is, 6 is equal to one of the direction angles of this 
normal. Therefore by Art. 209, 



cos 9 



dz 



Vr 

whence AP = --j- A*S r , 
dz 



* Of course z is also assumed to be real and continuous throughout S'. 
t The tangent planes must be so drawn that each face of P has the limit 0. 



330 INTEGRAL CALCULUS §216 



where 



MSMfHI)" 



a/7? 

Hence S = lim V AP = lim V ~ AS', 

17 IF d f 



and finally, 



-s S' 

dz 



x=S |/=/3(x) 




~te J J dz 

x=y y=a(x) 

the limits of integration depending upon the contour C or C. 
When the number of faces of P has become very great, each polyg- 
onal face, AP, coincides very nearly with the surface S itself, and 
may be regarded as a small portion of that surface. That is, 

Vp Vp 

-frjrdS , or -—rdxdy, 

dz dz 

is looked upon as the area of a small portion of the surface S, and 
is termed the element of surface. 

If, as in Art. 199, we use as the direction cosines 

— V —o 1 



Vl+p 2 + g 2 ' Vl+p 2 + g 2 ' Vl+pt + q*' 
forniula (C) becomes 

x=5 v=p(x) 

(CO S=f fVl+p* + q* 



x=S y=P(x) 

„ dx. 

x=y y=a(x) 



In this case the element of surface is V 1 + p 2 + g 2 cfo cfo/. 

By projecting $ upon the ?/2-plane or the £2-plane, we get S in 
the form 

\^dS'= f f—dydz, 



J J dx J J dx 

S' 2=7 V=a(z) 



§217 
and 

(C 2 ) 



s = 



MULTIPLE INTEGRALS 



2=5 z=/9(z) 

Vr 



331 



firii 






dxdz. 



S' z=y x=a(.z) 

S' denotes a different region, of course, in each of the three for- 
mulae (C), (Ci), (C 2 ). 

Example. Let us find by this 
method the surface of the sphere 
x 2 + y 2 + z 2 = a 2 . We have 

dx 



dz 



dy 
2 0, # = 4a 2 . 



Then 




VR = a = a 

df z Va 2 — x 2 — y 2 ' 
dz 

If S be the area of that part of the surface which lies in the first octant, 
we have 

x=a y=V fl 2-x2 x =a y^^at-z* 

dx 



x=0 



wa z 



S = af f — =JL dx = a r[sin-i -=¥=1 

J J Va 2 -x 2 -y 2 J [ Va 2 - x 2 \ 

z=0 y=0 

= I (sin- 1 l)dx — ~o I dx = 

2 = 

Hence the total surface = 8 # = 4 ira 2 . 

217. Exercises. 

1. Find the area of each portion of the surface of the sphere 
x 2 + y 2 + z 2 = a 2 
which is cut out by the cylinder x 2 + y 2 — ax = 0. 

Hint. S 



a ( sin -1 V _ 



+ * 



cfc. To integrate, set x = a tan 2 0. 



2. Find the area of the surface of the cone x 2 + y 2 — 2 2 tan 2 a = 
which is cut out by the cylinder (x — c) 2 -\-y 2 = a 2 , (a is constant.) 



332 INTEGRAL CALCULUS §218 

3. By the method of Art. 216, find the area of that portion of the 
surface of the cylinder ya -j- z* = a* which is contained in the cylinder 

2.2 2 

X3 +2/3 = a*. 

4. Use the method of this article to find the surface of the cylinder 
xa -+- 2/3 = as which is included in the sphere x 2 + y 2 + z 2 = a 2 . 

5. Find the surface of the sphere of 4 which is included in the cylinder. 

218. The Triple Integral. Let 7 be a portion of space com- 
pletely inclosed by a surface or by parts of several surfaces. Such 
is the portion of space inclosed by the surface of an ellipsoid, or 
by the four faces of a tetrahedron. Let f(x, y, z) be a function 
of three arguments, continuous at every point of V. Let V be 
cut up in any way into small elements of volume AV, and let 
(x, y, z) be any point in the interior of the element A V. Form 
all such products as f(x, y,z)AV and take their sum^/(a;, y, z)AV. 
Now let the number of elements of volume AF be indefinitely 
increased, but in such a way that the greatest diameter and con- 
sequently the volume of each AV shall have the limit 0. Then 
the fundamental theorem of the Integral Calculus of Functions 
of Three Arguments is 

jj£,f(x>9 y, z)AV has a determinate limit, and this limit is 
v 
independent of the manner of subdividing V into elements of 

volume. 

Tlris limit is defined to be the triple integral of f(x, y, z) extended 

over the volume V, and is denoted by the symbol / / / f(%, V, z) dvy 
so that v 

lim2)/(s, y, z)AV=ffff(x, y, z)dv. 
v V 

Moreover, the numerical value of the triple integral can always be 
found by three successive integrations. 

The following argument may serve to render probable the truth 
of the fundamental theorem, but is not to be taken as a rigorous 
proof of it. Let f(x, y, z) be the density of the element of volume 



§218 



MULTIPLE INTEGRALS 



333 



AT at the point (x,y,z). Then f(x, y, z)AV is approximately 
the mass of AV, and hence we may be reasonably certain that 

lim X/(z, y, z) AV = mass of V. 

v 

But the mass of V is constant, and therefore the fundamental 
theorem is true. 

We do not give a rigorous proof of the theorem. 

We shall prove, however, that the value of the triple integral 
can be found by three successive integrations. 

To this end we take, as elements of volume AV, the small 
parallelopipeds made by planes parallel to the coordinate planes. 
Let the distances apart of these planes be Azi, Az 2 , . . . , Ay h 
Az/ 2 , • • • , A.ri, Ax 2 , .... Then the elements of volume will 
have the dimensions Az, Ay, Ax, and the products will be of the 
form/(:c, y, z)AzAy Ax. Since the limit of the sum of products 
is independent of the manner of subdividing V, we may write 

J J J ^ x ' Vi ** dV ~ lim X^> V> *> Az Ay Ax ' 

v v 

Consider now that slice of V included between the planes x = x k 
and x = x k + Ax k . The section of V made by x = x k is shown in 
the figure and is denoted by R. The products belonging to this 
slice are 

X/(^> V> *) Ai2 ^ ^ Xk = kzk^Kxk, y, z) Az Ay, 



where x k and Ax k are constant. 
Now it is geometrically evident 
that 

lim^/fo y, z) Az Ay Ax = 






lim 2) Ax k Tlim ^f(x k) y, z) Az Ay~\. 

By Art. 214 we know that the 
limit within the brackets is the 
double integral of f(x k , y, z) (x k a constant) extended over R 




334 INTEGRAL CALCULUS §218 

(or over the projection of R in the yz-plane). We may write 
then 



\un£f(x k , y, z) Az Ay -ffffa, y, z) dzdy 

R R 

y=8(xk) z=\p(xk,y) 

J J ffrk, y, z) dz dy = F{x k ) . 



V=y(xk) z=4>(x k ,y) 

Jr> z =ip(x k ,y) 
f(xk, y, z) dz extends the summation along 
z=<f>(x k ,y) 

the line CD from z = 4>(x k , y) at C to z = \p(x k , y) at D. The 
second integration extends the summation over the whole of R, 
from y = y(x k ) at M or M i to y = 5(x k ) at N or Ni. We have 
now 

limV/(a;, y, z)AzAy Ax = \im^F(x)Ax = f F(x)dx. 

\r \r Ux=a 

Hence finally, 

(D) lim^fix, y, z) Az Ay Ax = j J J f(x, y, z) dz dy dx 
v v 

x=b y=8(x) z=>p(x,y) 

= f f f f(x,y,z)dzdydx. Q. E. D. 

x=a y=y(x) z=<t>(x,y) 

By cutting V into slices by planes parallel to the rcz-plane or the 
xy-pl&ne, it can be shown that it is quite immaterial in which 
order the three successive integrations are performed, care being 
taken in each case to employ proper limits. 

Example. Let us find the value of the triple integral J J j xdzdy dx, 

v 

where V is the space included in an octant of the sphere x 2 + y 1 + z 1 = r 2 . 
First Solution. We cut V into slices by planes parallel to the yz-p\sme. 
Calling the triple integral I, we have 



x=r y= v r 2 — x 2 z= V? i —x' i —y 2 

I = j J j x dz dy dx, 

x=0 y=0 z=0 

the limits of integration being determined as follows: 
In the first integration j x dz, we hold x and y constant and extend the 



§219 



MULTIPLE INTEGRALS 



335 



summation along the prism AB from 
z = at A to z = Vr 2 -x 2 -y 2 at B. 
In the second integration we hold x 
constant (z does not occur in the in- 
tegrand) and thus extend the sum- 
mation over all the prisms in the 
slice MPBN from y = at M to y 
-a yV 2 — x 2 at JV. The final integra- 
tion extends the summation over all 
the slices of the octant from x = N 

at to x = r at C The numerical calculation is as follows; 




v^~. 



J xdz — x V? 



3 = 



Then by the table of integrals 

2/= V r 2_a;2 



V r 2 — X 2 



1/2^ 



?/ W — x 2 — z/ 2 -f- (r 2 — a; 2 ) sin -1 



Z/=0 



Finally, 



Vi 



J y =o 



(r 2 — z 2 ):c. 



7 . | J> _ ^ & _ _ x |> _ ^J. g . A 



ns. 



Second Solution. Cut the solid into slices by planes parallel to the 
:n/-plane. Let the student draw the figure and complete the solution. 



219. The Triple Integral in Polar Coordinates. The posi- 
tion of a point P may be given 
by three coordinates, p, 6, 4>, as 
shown in the figure. We may 
regard p as the radius of the 
sphere on which P lies and whose 
center is at 0. 6 is the longitude 
of P, and 4> is its co-latitude. 
These spherical coordinates, as 
they are called, are connected 
with the Cartesian coordinates 




336 



INTEGRAL CALCULUS 



§219 



of the point by the equations 

x = p sin cos 6, y = p sin <f> sin 0, z = p cos 0, 
as is readily seen from the figure. 
We seek to express the triple integral 



///■ 



f(x, y, z) dV 



in terms of p, 9, 0. To this end we divide V into small elements 
of volume in the following way: 

First, through OZ we pass 
planes whose angular distances 
apart are A0i ; A0 2 , • . ■ • 

Second, about OZ as axis and 
with as vertex we describe cir- 
cular cones whose vertical semi- 
angles differ by A0i, A0 2 , .... 
Third, about as center we 
describe concentric spheres whose 
radii differ by Api, Ap 2 , .... 
V is thus divided into small elements of volume AV, such as 
PQRSP'Q'R'S' in the figure. P is the point whose coordinates 

are p, 6, 4>. 

PS = p A0, PQ = p sin 0A0,* PP' = Ap, 

and the product of these three quantities is p 2 sin0A0 A0 Ap. 

Now it can be proved that 

A7 




lim 



1. 



p 2 sin0A(/>A0Ap 

(Let the student give the proof of this. Note that A V = OP'Q'R'S' 
— OPQRS.) Therefore by DuhamePs theorem, Art. 173, we may 
replace AT in the triple integral by this product. Hence, setting 
f(x, y, z) = F(p, 0, </>), we have 

f ffffr, V, «) dV = lim 2) F(p, 0, 0)p 2 sin A<f> Ad Ap 
= lim 2) ^(p, A 0) A0 A0 Ap, 



* Because PQ is an arc of a smaZZ circle of a sphere. 



§220 • MULTIPLE INTEGRALS 337 

where G = Fp 2 sin <£. The last member of this equation is the 
limit of such a sum of products as is contemplated in our definition 
of the triple integral, and in formula (D), Art. 218, and may there- 
fore be evaluated by three successive integrations. Hence 

(E) ffff( x > y> *) dV = fff F ( p > e > 0)p2 sin dct> d ° dp 

v v 

p = b 6=&(4>) 0=/*(0,p) 

- / / / Fp 2 sin <t>d<f>dd dp. 

p =a 0=y(<f>) <t>=M0,p) 

Of course the order in which the integrations are performed is 
quite immaterial. 

Problem. Use spherical coordinates to find the value of J J J x dz dydx, 

v 
when V is an octant of the sphere 

x 2 + y 2 + z 2 = r 2 . 

220. Volumes by Triple Integration. If in formulae (D) and 
(E) we set f(x, y, z) = 1, the triple integral becomes in each case 
the volume of V. We have therefore 

(DO V = C C CdV = f f fdzdydx 

v v 

x=b y=5(x) z=4>{x,y) 

— I I I dzdy dx. 

x=a y=y(x) z=4>(x,y) 

(EO V= C C CdV = C f f p 2 sin <j>d<f>dd dp 

v v 

p =b e=8(<t>) 4>=n(e, P ) 

= / / / p 2 sin <f> dcp dd dp. 



p =o e=y(<i>) <t>=\(e, P ) 



CHAPTER XXX 

CENTERS OF MASS 

221. Center of Mass of a Solid. Let V be the volume and 
M the mass of a material body. Let the body be divided into 
small elements of volume AFi, AF2, . . . , and let their masses 
be AM i, AM 2 , .... Let (x i} y i} z») be any point of the mass 
AM » and form the products x { AM», yi AM», z { AM». Form such 
products for each AM and take the sums ^xAM, ^yAM, V^ AM, 
and 2 AM, the summations to extend over the entire body V. 
Now let the number of elements of volume (or mass) be indefinitely 
increased. We define the center of mass of V to be the point 
whose coordinates are 

lim ^x AM lim ^y AM lim ]£ z AM 
lim 2) AM ' lim 2) AM ' lim 2) AM ' 

These limits are plainly definite integrals, and if we denote the 
coordinates of the center of mass by x, y, z, we have 

Jx dM I xdM I y dM I y dM 

jdM jdM 



f/ dM ft 



zdM 
Jv J\ 

z = 



M 

dM 



Although we employ here but a single integral sign / , the inte- 
grals themselves may be single, double, or triple integrals, depend- 
ing upon the manner of dividing the body into elements of mass 
and upon the nature of the body. 

338 



§221 



CENTERS OF MASS 



339 



We shall not enter into the physical significance of the center 
of mass except to say that it is proved in books on mechanics 
that under the action of any system of forces a body will behave 
as though its mass were concentrated at the center of mass. 

Let mi> M2, • ■ • be the (average) densities of AM 1, AM 2 , . . . ; 
then AMi = tnkVi and equations (A) become 



(A') 



Lxdv r 



nydV 



i"> v 1/ 



a. 



ixzdV 



dV 



J,' 



dV 



fx may be a function of x i} yi, z i} or may be constant. In the 
latter case the body is homogeneous, and /* may be canceled in 
the fractions of (A'). We have then for a homogeneous body 



(A") 



X = 



CxdV CydV fzdV 

•Jy — ^V - V 

-,y = — tt-,z = 



v 



v 



In the case of a homogeneous body the center of mass is also called 
the centroid of the body, or of the geometrical figure. It will be 
observed that the integrals in (A r ) and (A") are special cases of 
(D), Art. 218. 



Example 1. To find the center of mass of a hemisphere in which the 
density is proportional to the distance from the base. 

Place the hemisphere on the axes as 
shown in the figure. Then ^ = kz, 
where k is constant. It is evident that 
the center of mass lies in OZ so that 
x = y = 0. Moreover, 



CkzzdV Cz 2 dV 

Jy Jy 

CkzdV CzdV 

Jv J V 




As element of volume we take a thin slice parallel to the base. AH the 
particles in such a slice will have the same density. The volume of such 
an element of volume is 7r(r 2 — z 2 )Az and its mass is x(r 2 — z 2 ) kzAz. 



340 



INTEGRAL CALCULUS 



§221 



Hence 



z — 



15 



15 r ' 



r / r 2 z 3 z 5\r 

„k fy-*)*dz - ( T - j =) 

Therefore the center of mass is at the point (0, 0, tzt). 



Example 2. To find the center of mass of an octant of the sphere 
x 2 -\- y 2 -\- z 2 = r 2 when the density is proportional to the distance from 
the yz-plane. 

In this case n = kx, and we have 



x = 



Cx 2 dV CxydV Cxz 

Jy _ Jy _ Jy 



dV 



jxdV jxdV jxdVi 




I Fig.l .Fig.2 

J 
Determining limits according to Fig. 1, we have 

x=r y=Vr 2 -x i z=-\/r 2 -x 2 -2/ 2 x=r y=Vr' i -x 1 

Cx 2 dV = j j fx 2 dz dy dx = fx 2 fVr 2 - x 2 - 



x=0 y=0 2=0 



X=0 2/=0 



y 2 dydx. 



J/ =Vr J -x* 



fVr 2 -x 2 -y 2 dy= ~[ y Vr 2 -x 2 -y 2 +(r 2 -x 2 )sm- 1 —^==\ = ^(r 2 -x 2 ). 
/-o L v =o ~ 



§222 



CENTERS OF MASS 



341 



To find CxydV it will be convenient to change the order of integration 
and to proceed according to Fig. 2. We have 



VV 2 -2 2 x=Vr i -y 2 -z 2 



y= Vr*-i 



l )ydydz 



15" 

.5 



fxydV-f f fyxdxdydz = lf j(, 

z=0 y=0 x=0 2=0 y = 

From symmetry it is evident that CxzdV = CxydV = '—. 

Jv Jv 15 

• As for the denominators in the expressions for x, y, z, we found in 

Art. 218, example, that | xdV= —• 
Jy 16 

pp, , - 7rr 5 Trr 4 8 - - a: 5 ^r 4 16r/ , ' 1 \ 
Therefore x = -^- = lT ; 2/=, = -^- = —^about-rj. 

Hence the center of mass is at the point ( — r, — - , -— ) or ( -— r, -r. ~r ) • 

\15 Iott lo7r/ \15 3 3/ 

222. Center of Mass of a Plane Area. The body whose 
center of mass we wish to find may be like a thin sheet of metal, 
whose thickness may be neglected, and 
we thus have the problem of finding 
the center of mass of a plane area. In 
such a case the volume V becomes the 
area A , and the element of volume A V 
becomes the element of area AA. If 
we suppose the surface to lie in the 
:n/-plane, z = 0, and we have 




(B) 



/ (xx dA I vy dA 

- Ja j - J A 

x= —n and v = ~r> 



j> 



dA 



j> 



dA 



Example. To find the center of mass of a quadrant of the area of the 
circle x 2 + y 2 = r 2 , when n = kx. 



In this case 



Cx 2 dA 

- Ja j - 

x = and y 

CxdA 



CxydA 
C xdA 



342 



INTEGRAL CALCULUS 



223 




We choose for element of area the strip 
shown in the figure. Then 



AA = y Ax = Vr 2 — x 2 Ax, 
and the element of mass is 

k Vr 2 — x 2 x Ax. 
Therefore 



f A x 2 dA= C Vr 2 — x 2 x 2 dx = r 4 f 2 sin 2 0cos 2 



dd 



irT< 



= (by the table of integrals) — - 

16 



Again, jxy dA = j(r 2 - x 2 )x dx = [~ —J = r -- 

Finally, CxdA = C Vr 2 -x 2 x dx=-\ \{r 2 - x 2 ) f\ * = - . 
J A Jo 3L J 3 



Therefore 



xr 4 r 3 Zirr , - ' r 4 r 3 3r 
16"3 = l6 and y= i^3 = V 



and the center of mass of the quadrant of area is at the point [ i = L ^, — ] 

V 16 4/' 

223. Center of Mass of an Arc of a Plane Curve. The body 
whose center of mass is sought 
may be of the form of a thin wire 
whose thickness may be neglected, 
and we get thus the problem of 
findinjg the center of mass of an arc 
of a/ curve (we here assume the 
curve to be plane). 

In this problem the volume V 
becomes the length of the arc, 5, and the element of volume be- 
comes the element of arc As. Hence 




(C) 



Jixxds I ju:r\/l + y'' 
s *J a 



dx 



J/mds f mv 1 

and a similar expression for y. 



+ y' 2 dx 



§224 



CENTERS OF MASS 



343 



Example. To find the center of mass 
of a quadrant of arc of the astroid 

2.2 2 

X 3-|- 2/3 = d3, 

supposing the density to be constant. 
We have ds = a^x* dx. 
Hence, 



x = y = 



xa*x *dx \x*dx -as rt 

•K = £* = 5_ 

I asz 3 da; j x *dx -as 



= 5 a - 




224. Center of Mass of a Surface of Revolution. The body 
may be a very thin shell in the shape 
of a surface of revolution. We neglect 
the thickness of the shell, and speak of 
X finding the center of mass of a surface 
of revolution. If the axis of revolution 
be OX, the element of volume becomes 
the element of surface generated by 
revolving about OX an element of arc As. That is, 

A 7 = 2^ As.* 

If we consider that case only in which the density is either 
constant or a function of x alone, we have y = z = 0, and 




(D) 



J-»x=& nx = b 
fixy ds I fixy V 1 + 
x=a U x = a 

2 d 



dx 



nyds 



f X fiyVl+y'*dx 

U x=a 



Example. Let the parabola y 2 = 4 ax be revolved about OX; to 
find the center of mass of that portion of the surface generated, which is 
cut off by the plane i = 3a, the density being constant. 



We have 



ds 



=\/ : 



x + a 



dx, yds = 2 \ r a^/x + a dx. 



* This equation is only approximately true, and the derivation of the for- 
mula? in the text is not rigorous, although the formulae themselves are accurate. 



344 

Hence 



INTEGRAL CALCULUS 



y/x + axdx r. _ , , , . 3 ~| 3 ° 



225 



Vx + adx 

n 



[(* + o)«]; 



58 
35 



a. 



Problem 1. Obtain a formula for the center of mass of a solid of revo- 
lution. 

Problem 2. Devise a rigorous demonstration of formulae (D). 

225. Center of Mass of Any Curved Surface. When the body 
whose center of mass is sought is a thin curved shell, whose thick- 
ness is so small that the shell may be regarded as a curved sur- 
face, the element of volume becomes the element of such a surface, 
and is, by Art. 216, 

-Tjfdxdy, 

~dz 



Therefore 



where 




Vr. Vr 

—tt dy dz, or —rj dz dx. 
nyVR 



dx 



i 



dz 



dxdy 



V = 



M 



M = 



a 



tx Vr 

dz 



dxdy. 



By using one of the other elements of 
surface, we get other similar expressions 
for x, y, z. 

Example. To find the center of mass of that 
portion of the surface of the homogeneous sphere 
■X x 2 + y 2 + z 2 = a 2 

which lies above the x?/-plane and is cut out by 
the cylinder 

x 2 + y 2 — ax = 0. 



2 denotes the entire circular base of the cylinder. 



§225 CENTERS OF MASS 345 

It is readily found that 

VR _ a = g 

Jj_~ z Va?-x 2 -y 2 
dZ 
Hence. 



U 



Vc 



dxdy 



S 

x—a y=^ax—x 2 

Numerator = 2 a \ x I dx 

J J V a 2 - x 2 - y 1 

i=0 y=0 

£=a y= Vax—x 2 

= 2 a |x sin -1 = \ dx = 2a I x sin -1 v — % — dx, 
J L Va 2 -x 2 ] Jo a + x 

3=0 2/=0 

This last integral may be evaluated by setting x = a tan 2 0. We get 
finally, 

Numerator of x = f a 3 . 

In exercise 1 of Art. 217 it is found that S, the area of that portion of 
the spherical surface whose center of mass is sought, has the value 
a 2 (7r - 2). Therefore 

x 
We have further 



Z = 



fa 3 


2 « _.__ r> rOA n 


a 2 (7r-2) 


3( ,r_2) 




x=a j/= y/ax— x 3 


J J adydx 


2 a J J dydx 


s 


x=0 y=0 


S 


£ 



Numerator = 2 a J Vaz — x 1 dx — ~' 



Therefore 



x=0 

-n-a 3 

4 ttO 



a 2 (7r-2) 4(tt-2) 



0.688 a. 



From the symmetry of the figure it is evident that y = 0. Therefore 
the center of mass of the surface in question is at the point 
(0.584 a, 0, 0.688 a). 



346 INTEGRAL CALCULUS §226 

226. Exercises.* 

Find centers of mass of the following bodies. Unless the contrary is 
stated, the bodies are assumed to be homogeneous. 

1. The volume of an octant of a sphere. 

2. The surface of a hemisphere when the density varies as the distance 
from the base. 

X% y2 tf 

3. That half of the volume of the ellipsoid — + 7^ H — = 1, which 

a 2 b 2 c 2 

lies at the right of the ?/2-plane. 

4. The volume of exercise 3, when n = kx. 

5. An octant of the volume of the ellipsoid of exercise 3. 

6. A quadrant of area of the ellipse — + 7- = 1. 

a 1 b l 

7. The area bounded by the upper branch of the parabola y 2 = 2 mx, 
the x-axis, and the line x = a. 

8. The volume generated by the revolution about OX of the area of 
exercise 7. 

9. A straight wire whose density is proportional (1), to sin^y- ; (2), to 

sin^y, I being the lengths of the wire, and x being measured from one 

end. 

10. The area of a plane triangle. 

11. The volume of a regular tetrahedron. 

12. A quadrant of area of the astroid x* + 7/3 = a*. 

13. The volume generated by revolving about OX the area of exer- 
cise 12. 

14. The area of the first arch of the cycloid 

x = a(d — sin d), y = ail — cos 6). 

15/ The arc of the first arch of the cycloid. 

16. The area of the cardioid p = 2a(l — cos 9). 

17. The arc of the cardioid. 

18. The volume generated by revolving the cardioid about the initial 
line. 

19. One-half of the area of the lemniscate p 2 = a 2 cos 2 d. 

20. The convex surface of a right circular cone. 

* For an elementary treatment of Centers of Mass and of kindred subjects 
the reader is referred to two excellent little books, one entitled Applications 
of the Calculus to Mechanics by Hedrick and Kellogg, and the other The In- 
tegrals of Mechanics by Lester, both published by Ginn & Co. A number of 
the exercises of this article are taken from these books. 



§226 CENTERS OF MASS 347 

21. A sector of a circle of central angle 2 a. Use polar coordinates, 

and the bisector of the sector for initial line. Consider the case « = 7 • 

4 

22. An octant of volume of a sphere whose density is proportional to 
the distance from the center. Use spherical coordinates. 

23. Given the catenary y = -\e a -{- e a ) , and two points of the curve 

z 

whose abscissas are — c and +c. Let k be the common ordinate of these 
two points, and 2 I the length of the arc between them. Express the 
coordinates of the center of mass of this arc in terms of a, c, Jc, and I. 

24. Solve exercise 23 on the supposition that n = bs, s being measured 
in both directions from the y-&xis. 

25. The surface of an octant of a sphere. 

26. The surface of a quadrant of a sphere. 

27. That portion of the surface of the sphere x 2 + y 2 -f- z 2 = a 2 which 
lies in the first octant and within the cylinder x* + y* = a*. 

28. The portion of the surface of the cylinder in exercise 27. 

29. Prove the following theorem, which is termed 
The First Theorem of Pappus. 

The volume of the ring generated by the revolution of a plane area A 
about an external axis in its plane is equal to the volume of a right cylinder 
whose base is A and whose altitude is the distance traversed by the 
center of mass of A, A being homogeneous. 

30. Prove the following theorem, which is known as 
The Second Theorem of Pappus. 

The surface of the ring of exercise 29 is equal to the convex surface of 
the right cylinder. 

N. B. The principles proved in exercise 14, Art. 177, are special 
cases of these two theorems of Pappus. 

31. When the curve that bounds an area is given in polar coordinates, 
show that the coordinates of the center of mass of the area are given by 
the formulae 

f \ p 2 cos ddpdd - \ p 3 cos Odd - J p 3 sin dd 



x = = — , y = 



ffpdpdd \jp 2dd \j p2dd 

32. Use the results of exercises 31 and 29 to show that the volume 
of a solid of revolution about OX, when the generating curve is given 
in polar coordinates, is given by the formula 

2 r 

V = - 7T J p 3 sin d dd. 



BOOK V 

THEOREMS OF TAYLOR AND MACLAURIN. 
INTEGRATION OF RATIONAL FRAC- 
TIONS. ENVELOPES 






CHAPTER XXXI 
THEOREMS OF TAYLOR AND MACLAURIN 

De Moivre's Theorem and the Hyperbolic Functions 

227. Expression of a Function as a Polynomial. We repeat 
here with slight variation the demonstration of the Law of the 
Mean given in Art. 114. We set 

(a) f(b)=f(a)+K(b-a), 

and shall determine a form for K by means of Rolle's theorem. 
From (a) we get 

(a') f(b)-f(a)-K(b-a) = 0. 

We now form a new function, 4>(x), by writing x for a in (a'),* 
<t,(x)=f(b\-f(x)-K(b-x). 

Differentiating this, 

</>'(*)= -f'(x) + K. 

Now <f>(x) and <t>'(x) are continuous throughout the interval 
a = x = b, because f(x) and f(x) are so by hypothesis. More- 
over, (f>(a) = by (a') and 0(6) = 0, identically. Hence Rolle's 
theorem applies to <j>(x), and therefore 

0'[o + 6(b - a)] = -{'[a + 0(6 - a)] + K = 0, 

where < 6 < 1 ; 

* .EC is at present unknown and may of course be a function of a, but, in 
forming the function <f>(x), the a that may occur in K is not changed to x. 

349 



350 CALCULUS § 227 

whence 

K=f[a + 6(b-a)]. 

Substituting in (a), we have 

(1) f(J>)=f(a)+f'[a + 0Q>- a)](b - a), 

which is the Law of the Mean (see (a), Art. 114). 

We shall now obtain by the same method an expression for 
f(b) that shall contain b — a and (b — a) 2 . We assume that 
f"(x) as well as }{x) and/'(z) is continuous throughout the interval 
a = x = b, set 

(b) f(b) = /(a) + /'(a) (6 - a) + K(b - a)\ 

and shall determine a form for K by the aid of Rolle's theorem. 
From (b) follows 

(b') Kb) - f(a) - f(a) (b-a)- K(b - a) 2 = 0. 

We form a new function, 4>(x), by writing x for a in (b r ),* 

4>(x) =/(&) - f(x) - f'(x) (b-x)- K(b - x)\ 
Differentiating this 

4>'{x) = -f'(x)+f(x) -f"(x) (b-x) + 2K(b - x) 
= (b-x){-f"(x) + 2K}. 

Now 4>(x) and <f>'(x) are continuous throughout the prescribed 
interval, because f(x), f(x), and }"{x) are so. Moreover, </> (a) = 
by (b'), and 4>(b)=0 identically. Therefore Rolle's theorem 
applies to 4>(x), and we have 

4> f [6+d{b-a)]=0 or (1-0) (b-a) { -f"[a + &Q>-a)]+2K\ = 0. 

But neither 1 — 6 nor b — a are 0, and therefore the factor within 
the \ | must be 0. Hence 

f"[a + 6(b - a)] 
K = 2 

Substituting in (b), we get 

(2) Kb) = /(a) + /'(a) (6 - a) + /> + ^ ~ °)1 (6 _ a)2 . 

* See footnote on page 349. 



§227 THEOREMS OF TAYLOR AND MACLAURIN 351 

Let us carry this process one step farther. We assume that 
f(x), f'(x), f"(x), and f'"(x) are continuous throughout the pre- 
scribed interval, and set 

(c) f(b) = /(a) +f>(a)(b-a)+Hf(b-ay + K(b- a)\ 

From (c) we get 

(c') Kb) - /(a) - /'(a) (b - a) - M (6 - a) • - K(b - a)» = 0. 

We form now a new function, <f>(x), by writing x for a in (c')>* 

*(*) = /(&) - /(*) - /'(x) (6 - x) - £^ (6 - x)« - JT(6 - x)K 
Differentiating this, 

-^-^■(b - z) 2 + 3i£(6 - z) 2 
( f'"(r) 

Here too 0(x) is seen to fulfill the conditions of Rolle's theorem, 
and therefore by that theorem 

4>'[a + 0(6 -a)] = 0, 

or (i-enb- a y\- f '" [a + e 2 {b - a)] + 3K\ = o. 

Consequently 

j. f'"[a + 9(6 - a)] . 

K = 3! 

Substituting this value for K in (c), there results 

(3) /(6)=/(a)+/'(a)(6-a)+ffl(6-a) a 

+ S'"la + W-*)]\ b _ a)K 

Let the student show, by the same method, that 

(4) S(b) = S(a) +/'(o) (b - a) + f -^ (b - a) 2 + ^ (6 - a)' 

+ /I y [a + y_ g)] (6 _ g)4 

* See footnote on page 349. 



352 CALCULUS §227 

It is scarcely necessary to point out that the 0's of formulae 
(1) . . . (4) are not ordinarily the same. 
The important fact to observe is that in each of these formulae 

(1) . . . (4) we have f(b) expressed as a polynomial in (b — a)* 

And it is obvious that by the same process may be obtained an 
expression for /(&) that shall contain as many powers of (b — a) as 
we please, provided always that f(x) and as many of its derivatives 
as appear in the polynomial are continuous throughout the pre- 
scribed interval. 

On making a slight change in the notation by writing b = a + h, 
and b — a = h, formulae (1) . . . (4) take the forms 

(1') f(a + h)=f(fl)+f'(a + 6h)h. 

(2') f(a + h) - f(a)+f'(a)h + f " (a + 6h) h*. 

(30 /(a +h)= m+rm +^r h2 + / ' (a 3 | 0h) h *- 

(40 /( „ + k) = /(a) +na)k +«A° +«* + f ^±°»k>. 

Each of these formulae expresses f(a + h) as a polynomial in h* 
And the law of their formation becomes evident on inspection, so 
that successive polynomials can be written down at will. Each 
polynomial contains, of course, an unknown element 6. 

The advantage of putting a function into the form of a poly- 
nomial is that a polynomial is the simplest kind of a function to 
handle. For instance, its numerical value can be calculated for 
any value of the argument. 

To illustrate the application of the foregoing formulae, let us 
express sin (a + h) as a polynomial in h. In this case we have 

f(x) = sinz, }'{x) = cosx, f"{x) = —sin a;, f'"(x) = —cos a;, 
/ IV (z) = sinz, f v (x) = cos a;, etc. 

* In calling the second members of (1) . . . (4) and of (1') . . . (4') polyno- 
mials in b — a and in h, we leave out of consideration the b — a and the h 
that occurs in the derivative in their final terms. In strict sense they are not 
polynomials, unless f(x) is itself a polynomial. 



§227 THEOREMS OF TAYLOR AND MACLAURIN 353 

Plainly, f(x) and all its derivatives are continuous for all values of 
x. Substituting, therefore, in formula? (1/). . . . (4'), we have 

(Si) sin (a + h) = sin a + cos (a + 6h) • h. 

/a \ • / i i^ ■ i z, sin(a + 0/i) 

(5 2 ) sm (a + /i) = sin a + cos a • A = h\ 

/a \ • / i n • i i, sin a,, cos (a + 0ft) ,, 

(53) sin (a + /i) = sin a + cos a • /i ^- /r ^ /V\ 

(5 4 ) sin (a + h) = sin a + cos a* h ~p ft 2 ~j— A 3 

, sin (a + 0/Q M 
f 4 , " • 

Additional formulae can be written down at will. An inter- 
esting special case is gotten by setting a = in the foregoing 
formula?. Then sin a = 0, and cos a = 1, and we have 

(Si') sin ft = cos 0ft • ft. 

fa n -li. sin 0ft , 2 

(5 2 ) smft = ft ^ — ^ ■ 

(5 3 ) sm ft = ft ^7— ft 3 . 

o! 

/d /\ • 7 7 ^ I Sm ^ 7 A 

(S/) sin/i = h - gj .+ -^- ft 4 . 

, a ,. . , , ft 3 cos 0ft , 
(S 5 ') sm ft = ft - ^ H ^— /i 5 . 

(5 6 ) sin A = ft — g| + ^ gp A 6 . 

/a ,n • 7 7 h 3 ft 5 cos 0>i , 7 

(5 7 ) sin ft = ft — ■ gj + gj ^— A. 

/an ■ z. 7. ^ 3 1 ^ W sin 0ft 

(5 8 ) smft = ft - -, + - ; - ^ + ~$i~ h • 

These formula? enable us to compute the numerical value of 
sin/i for any given value of ft, but only approximately because 
of the presence of the unknown 0. However, the approximation 



354 CALCULUS §227 

can be made very close. Thus, if we neglect the last term in 
formula (S 6 ')> and write 

the error committed will be less than ^-. , a number that is very small 
when h is small. For example, suppose h = - = 20°. Then 



sin 



ono • 7T 7T 1 /ir\ 3 1 Ar 

20 = Sln - = ___y + s ! fg | 



which is too large (because sin 6h is positive) by a number less than 
By means of a table of logarithms we find that 



6<W- 



Hence 



7T 

9 ~ 


.349 066, 


(^Y= .042 533, 


r 


.005 183, 


(|Y= .001809. 




7T 

9~ 


.349 066 




3!W 


-.007 088 




5!W 


.000 043 



| sum = .342 021 

Therefore sin 20° = .342 021, a result which is too large by a 

number less than^rf-J = .000 0025. We have then a very close 

approximation to the true value of sin 20°. And by using a formula 
that contains a large enough number of terms, we can compute 
sin h for any value of h to any degree of accuracy. 

It appears, then, that one of the advantages of being able to 
write a function in the form of a polynomial is that by this means 
we may calculate the numerical values of the function for given 
values of the argument. 






§§228-229 THEOREMS OF TAYLOR AND MACLAURIN 355 

228. Exercises. 

1. Express cos (a + h) as a polynomial of the sixth degree in h. Place 
a = in this result, and thus express cos Hsa polynomial in h. Com- 
pare this expression for cos h with the result of differentiating (S 7 ') of 
Art. 227. 

2. Express e a+/l as a polynomial of the sixth degree in h. Place a = 
in this result, and thus express e h as a polynomial in h. In this last 
result set h = 1, and compare with equation (2) of Art. 68. 

3. Express log (a + h) as a polynomial of the eighth degree in &. Place 
a = 1 in this result. 

4. Compute sin 30° and sin 36° correct to 5 places of decimals. 

229. Taylor's Theorem: Finite Form. Taylor's theorem is 
merely the general formula of which (1) . . . (4) or (1') . . . (4') 
of the preceding pages are special cases, and its form could be 
inferred from an inspection of those formulae. In giving here the 
complete proof of Taylor's theorem, we shall obtain two forms for 
the final term. 

Proof of Taylor's Theorem. We assume that f(x) and its first n + 1 
derivatives are continuous* throughout the interval a = x = b. 
We set 

(a) /(&) = f(a) + f(a) (6 - a) + ^ (6 - a)" + Q^ (b - a)' 
+ ■ ■ ■ +\^^ l ( b - a ) n - 1 + J ~r(.b-a)»+K(b-a)>, 

the form of this expression being suggested by formulae (1) . . . 
(4^ of Art 227 No restriction is placed upon s except that it 
shall not be nor oo . From (a) follows 

(a') f(b) - f(a) - f'ia) (6 - a) - f -^ (6 - a)» - . . 
ft"- 1 ) (a) {("Ha) 

- fe^iyf (6 - a)n ~ l ~ m {b ~ a)n ~ K(b ~ a) ' = °- 

* Here as elsewhere it is to be understood that the functions are assumed 
to be single-valued and differentiable, as well as continuous. 



356 CALCULUS § 229 

We form now a new function, 4>(x), by writing x for a in (a'),* 

3! i0 ; ' * " (n-1)! l ; 

. - l — p- (b - x) n - K(b - x)\ 

Differentiating this equation as to x, we have 




4>'(x) = (b- x) s ~ l j - f {n+1 \ (x) (6 - x) n ~ s+l + si£ J • 

<f>(x) and <£'(#) are continuous throughout the given interval 
because f(x), f(x), . . . / (n+1) (a;) are so. Moreover, 4>{a)= 0, by 
virtue of (a'), and <j>(b) = identically. Hence Rolle's theorem 
applies to <f>(x) } and we have 

4>' [a + 0(6 - a)] = 0, 

or (i - ey-'ib-ay^x 

{ - (1 - g) - +1(& - tt) .-. + x / ( " + »[" + »<» - «>1 + sg j , o. 

Since neither (1 — 0) s_1 nor (& — a) s_1 are 0, the quantity within 
the { I must be 0. Therefore 

K _ f^na + eib-a)] _ _ 

s- ft! 

and the last term in (a) takes the form 

_ = /^[q+^-q)] _ 6 _ fl)n+1 _ 

s • n\ 
Now we may assign to s any value we choose except or oo, 
The most useful values are n + 1 and 1. 

* See footnote on page 349. 



§229 THEOREMS OF TAYLOR AND MACLAURIN 357 

When . = » + 1, K{b - a)> J {n+1)[ « + e t- a)] (b - «)-h. 

in + 1) ! 

When s = 1, K(b — a) s = - — l - — — - - (b — a) n+1 . 

Therefore finally, 

(A) /(6)=/(a)+/'(a)(6 - a) + f ^ (b - a)« 

where fi= /^>[a + y- a )] 

(n + 1) ! 

(1 - 9)V(B+1)[ + e ( & _ a)] 

71 ! 

Formula (A) is Taylor's theorem, or Taylor's formula. R is termed 
the remainder in Taylor's formula. Formulae (1) ... (4) of 
Art. 227 are plainly special cases of (A) corresponding to n = 0, 
1, 2, 3. 

The more convenient form of Taylor's theorem is obtained by 
setting b — a = h (whence b = a + h) in (A) , and is 

(B) /(« + ft)=/(a)+/'(o)A+£^fc' 

wta,g- ^7<« + *> *•« or (l-WCa + Sfry, 

(n + 1)! n ! 

Formulas (1') . . . (4') of Art. 227 are special cases of (B). There 
is of course one element of uncertainty in Taylor's theorem, viz., 
6, concerning which we know only that < < 1. The second 
member of (B) is a polynomial of degree n + 1 in h (leaving out of 
account the h that occurs in/( n+1) (a + Oh)). This expression for 
/(a + h) is termed the development of f(a + h) in powers of h, 
and in applying Taylor's theorem we are said to develop the 
function. 



358 



CALCULUS 



§230 



Y 








c 


)S 








' T 






P^^ 







B 







Aa) 


h 






c 


i 




a- 


Yh 



In the figure 

BQ = f(a + h), BC = /(a), CT = f'(a) - ft. 

Hence TQ represents in magni- 
tude and sign the sum of all the 
terms after the second in the 
second member of (B), and that 
too for any value of n. 

In formula (B), h is the incre- 
ment of a, and by transposing 
/(a) to the first member we get 

/(a -f- h) — S(°) = 4f ( a ) • If now we write x for a and dx for h, 

and observe that 

f(x)dx = df, f"(x)dx 2 = d 2 f, . . . f n (x)dx n = d n f, 

(B) takes the form 

(BO A /=c?/ + ^ + i^/+. • .+±dTf + R, 

where # = -. v , ' , — -dz n+1 or - — ^ — L dx n+1 , 

{n + 1)! n\ 

a formula which expresses the increment of a function as a sum of 

multiples of its differentials. 

230. Examples of Development by Taylor's Theorem. 

1. To develop sin (a + h). 

We have already had, in Art. 227, developments of sin (a + h) 
for n = 0, 1, 2, 3, but with only one form of remainder. We 
shall now appty Taylor's theorem to obtain the general develop- 
ment with both remainders. We have 

fix) = sinx, f'(x) = cosz, f"(x) = — sinx, . . . 

f»(x) = sin (a; + »|Y /(» +1 >(z) = sin|z + (n + 1)||- 

Therefore, substituting in (B) , we have 

(S) sin (a + h) = sin a + cos a*h ^r- h 2 



cos a 
"3T 



2! 
sin 



A 3 + 



+ 



n(a + n|) 



ra! 



h n + R, 



§230 THEOREMS OF TAYLOR AND MACLAURIN 359 

where 



sin 



f"a+(n+l)|+^l (l-^)»Bin[a+(n+l)|+flfcJ 



R = — t — t , " *h n+1 or *= =^»+i. 

(w+1)! n! 

If in this formula we set n = 0, 1, 2, 3, we shall get of course the 
developments (Si). . . . (S 4 ) of Art. 227. 

If in (S) we set a = 0, there results (writing x for h) 



sin 



(!) 

Un + 1)|+ tej (1 - ^sniffa + 1)|+ ftrl 



/>»3 /v»5 o*7 /v»9 \ 

(SO sinaj^aj-gj + gj-yj + gi-. ■ •+— ^p 

where 



# = — *=—. — — p-. -x n+1 or ^—. -x n+1 . 

(w + 1)! n\ 

Formulae (S/) . . . (S 8 ') of Art. 227 are special cases of (S'). 
2. To develop log(a + /i). 
In this case 

. . . /»(,) . (-D-'(n-l)l and /( „ +I)W . (=!£«! . 

On substituting in (B) , we have 

(L) log(a + h)=Aoga + ^^ + ^ 

4 a 4 v J na n ' 



where 



(- l) n h n+1 (0- l) n h n+1 

(n + 1) (a + <9/0» +1 ° r (o + ^)» +l 



When a = 1, log a = 0, and we have as special case of (L) 
(writing x for h), 

/v«2 /v»3 /y»4 /*»5 /y»7l 

(L')\og(l+x) = x-~ + ^~+j +(_i)»-i£. + fi j 

where 

^ (-l)n^+l (fl _ l)n^+l 

(n + 1) (1 + BxY+ l ° r (1 + ^) n+1 * 



360 CALCULUS §§231-232 

231. Exercises. 

1. Write (S r ) of example 1 with both forms of R f or n = 7, 8, 9, 10. 
Put x = 1 in formula (L'), take n = 10, omit R, and get 
log 2 = .645 635. 
Show that this value is too small by a number that lies between — and 

— • — . These are wide limits of error. The true value is log 2 = 

l i z 

.693 147, so that (L') does not give accurate results unless n is taken very- 
large. A more useful formula for the calculation of logarithms is given 
on a later page. 

3. Obtain the general development of cos(a + h). 

4. In the result of 3 set a = 0, and h = x, and get 



cos 



HL 



where 

cos [(n + 1) ~ + dx~] (1 - e) n cos \{n + 1)| + Bx\ 

R = L , , i , i -x n + l or ±- f ^ ^x"+K 

(ft + 1)! n\ 

5. Obtain the general development of e a + h . 

6. In the result of 5, set a = 0, and h = x, and get 

/y»2 /y»3 /v»4 /y*^ 

CEO e » =1 + I + | + |. + | j+ ... +£.+«, 

where 

# = 7-^-TT,^ n+1 or ^ ^-a^ 1 . 

(n+1)! n\ 

232. Maclaurin's Theorem: Finite Form. This is a special 
case of Taylor's theorem obtained by placing a = in (B) of 
Art. 229. If at the same time h be changed to x, there results 
the customary form of Maclaurin's theorem, 

f"(0) f'"(0) 

(C) /(*) = /(0) +/'(0) x + J -^p x> + J -^ z 3 + 



where 



/w)(to) ^, or (1 -<>)"/("+'> (to) ^ 
(n + 1)! n! 



§§233-234 THEOREMS OF TAYLOR AND MACLAURIN 361 

Example. To develop sin x by Maclaurin's theorem. 
In example 1 of Art. 230, we have calculated the derivatives of sin x. 
Putting x = in these, we have 

/(0) = 0, /"(0) = 0, /*(0) = 0, /"((J) = 0, . . . , 
/'(0) = 1, /"'(0) = -1, /v(0) = 1, /vn (0) = -1, . . . , 

/(») (0) = sin (n |\ / (n+1) (to) = sin Yin + 1) ^ + to] • 

Substituting in (C), we get formula (S') of Art. 230. 

In the problems of the preceding pages where we have developed 
f(a + h) by Taylor's theorem, and have then set a = 0, the re- 
sults are precisely the same as would be obtained by developing 
the f(x) by Maclaurin's theorem. 

233. Exercises. Develop the following functions by Mac- 
laurin's theorem: 

1. cos a;. 2. e*. 3. xe x . 

4. log(l-r-z). 5. sin 2 a;. 6. cos 2 a:. 

7. Why cannot log x be developed by Maclaurin's theorem? 

8. f(x) = Co + Cix + c 2 x 2 + c z x z , the c's being constants. Develop 
/(a + h) by Taylor's theorem. 

9. Develop the polynomial of exercise 8 by Maclaurin's theorem. 
10. Develop (a + h) b by Taylor's theorem. 

234. A Property of Polynomials. Let P(x) be a polynomial 
of mth degree in x. Since P (w) (z) is a constant, all higher deriv- 
atives are 0, and consequently if we develop P{x) by Taylor's 
formula (A) and take n = m, R will be 0. Doing this, and writing 
x in place of b in (A) , we have 

P(x) = P(fl) + P'{a) (* - a) + ^J^ (x - a)» 

4- — gj- y (x - a) 3 + • • • + —^7- O - «) w - 

From this development it is readily seen that the necessary and 
sufficient conditions, 

that P(x) be divisible by (x — a) is P(a) = 0; 

that P(x) be divisible by (x - a) 2 are P(a) = 0, P'(a) = 0; 



362 CALCULUS §235 

that P(x) be divisible by (x — a) 3 are 

P(a) = 0, P'{a) = 0, P"(a) = 0; 

and in general 

that P(x) be divisible by (x — a) r are 

P{a) = 0, P» = 0, . . . PC'-D(o) = 0. 

We now define a to be an r-fold root of P(x) when P(x) is divisible 
by (x — a) r . We have then the 

Theorem. The necessary and sufficient conditions that a he an 
r-fold root of P(x) are 

P{a) = 0, P\a) = 0, P"(a) = 0, . . . PC- 1 ) (a) = 0. 

Problem 1. What is the geometric meaning of the theorem in the case 
of a double root? of a triple root? 

Problem 2. Develop P'{x), P"(x), . . . P(r~V(x) by Taylor's theorem 
and show that 

(a) A double root of P(x) is a simple root of P'ix); 

(b) A triple root of P(x) is a double root of P'ix), and a simple root of 
P"(x); 

(c) An r-fold root of P(x) is an (r — 1) fold root of P'(:r), an (r — 2) 
fold root of P"(x), ... a double root of P<- r - 2 )(:c), and a simple root 
of PC^Oc). 

235. Evaluation of the Indeterminate Form ^ by Taylor's 
Theorem. Suppose 

\!M~\ We wish to determine lim ^ • 

To make the problem general, we shall suppose that 

fix) fix) f'jx) f^-Vjx) 

Q(xY g'ixY g"ixY ' ' ' g<"-»(x) ' 

all take the form r when x = a. 

On developing both numerator and denominator of ~ — r-~ by 

gia + h) 

Taylor's theorem, there results, under the given hypotheses, 



§236 THEOREMS OF TAYLOR AND MACLAURIN 363 

f(a + h) n\ a ^ (n + 1)! 



Therefore 



g(a + h) flf(")(g) ^ n g^Co + g^) ^ n+1 
n! (n + 1)! 



which is equivalent to the rule of Art. 116. Of course it is here 
tacitly assumed that f(x), g{x), and their first n + 1 derivatives 
are continuous in the vicinity of x = a. 

236. Criteria for Maxima and Minima Determined by Taylors 
Theorem. 

Lemma 1. In the polynomial 

b Q + bih + b 2 h 2 + bji 3 + • • • + b k h\ 

h can be made so small that the polynomial shall have the sign of 
its first term bo. 

For, since the 6's are finite constants, and there is a finite number 
of them, h can be made so small that 

bih + b 2 h 2 + bsh 3 + • • • + b k h k = h(bi + b 2 h + • • - + b k h k ~ l ) 

shall be less than any quantity that can be named, and hence it 
can be made less than &o. And then the polynomial will have the 
sign of 6 - 
Lemma 2. In the polynomial 

boh m + bih m+l + b 2 h m+2 + • • • + b k h m+k , 

h can be made so small that the polynomial shall have the sign of its 
first term, boh m . 

For, the polynomial may be written in the form 

h m (bo + bih + b 2 h 2 + ■ ■ • +b k h k ), 

and by lemma 1, the polynomial within the parentheses can be made 



364 



CALCULUS 



§236 




to have the sign of 6 , by making h small enough, and then the 
given polynomial will have the sign of its first term, boh m . 

A study of the accompanying figures will make it plain that our 
definitions of maximum and minimum, Art. 59, are equivalent 
to the following 

Definitions. If f(a + h) — f(a) and f(a — h) — f(a) are both 

, j for all values of h within a 

sufficiently small interval, PQ, 
that contains x = a, then /(a) 

maximum 
minimum 

We have now to prove the 
following 

Theorem. Assuming that f(x), 
f'(x), f"(x), and f"'( x ) are con- 
tinuous throughout an interval 
that contains x = a, the necessary 
and sufficient conditions that f(a) 

( maximum 
I minimum 



is 



P a q 



Y 






.1 




^^Wl 









f(«> 


X 




i 


3 a <; 


) 



be a \"™^7™::\ are that /'(a) 



shall be and /"(a) 



+ 



First. To prove these condi- 
tions necessary; that is, to prove 

that if M « EZn 1 !>/>>= °> and '"W is j; 

We develop f(a + h) to four terms by Taylor's formula (B), or 
by (30 of Art. 227, and then transpose f(a) to the first member of 
the equation: the result is 

(a) f{a+h) - f(a) = rm+ m h ,+ r^^ h *. 



2! 



3! 



Writing —hiovh in this equation, 

(b) /(a - h) - f(a) = -f'W + f -^-h* - f " (a ~ BK) ». 

Since /(a) is |™^um ! by ^thesis, the first members of (a) 



§236 THEOREMS OF TAYLOR AND MACLAURIN 365 

and (b), and consequently their second members, must, by defi- 
nition, have the same sign for all values of h in the vicinity of 
x = a. But by lemma 2, for small values of h, the second members 
of (a) and (b) will have the signs of their first terms. Conse- 
quently their first terms cannot be /'(a) A and — f'(a)h, and therefore 
/'(a) must be 0, and the first condition is proved necessary. Since 

the first term of the second member of both (a) and (b) is . A 2 , 

h 2 . 
and since -^ is always +, the second members, by lemma 2, will 

have, for all small values of A, the sign of f"(a). Hence the first 
members of (a) and (b) will have the sign of /"(a) for small values 
of A. From this and from our definition it follows that if /(a) 

is Imfmnium \' ^"( a ) is + ' and ttlus tiie seconcl conca tion is 
proved necessary. 

Second. To prove these conditions sufficient; that is, to prove 

that if f'(a) = 0, and /"(a) is j ~ j , then /(a) is j ^hZm j ' 
Since f'(a) = 0, we have 

/(a-AW(a) = »*-&j^. 

By lemma 2, for sufficiently small values of h, the second members 
of these equations, and consequently their first members, will 

have the sign of their common first term,"^— ^ — A 2 , that is, the sign 

of /"(a). Therefore /(a) is either a maximum or a minimum. 
And if the sign of }"{a) and consequently of f(a + h) — }{a) and 



f(a — h) — f{a) is } , ( , it follows from our definition that f(a) is a 

mmTmum * ^^ e con ditions are thus shown to be sufficient. 

N. B. In the foregoing discussion it is tacitly assumed that 
/"(a) ^ 0. 




366 CALCULUS §237 

Problem. Assuming that f(x) and its first n + I derivatives are con- 
tinuous in the vicinity of x = a, employ Taylor's theorem to prove that 
if f(fl) , f (a),f'(a) , . . . /< s) (a) are all (s < n), and if s is an odd number, 

and if /(->(«) i S |;j,then/( a ) is a|-™!- 
What conclusion can be drawn if s is an even number? 

Because in deriving Taylor's 
formula (B) we assumed all de- 
rivatives to be continuous (and 
therefore finite), we cannot by 

this method determine the conditions for maxima and minima of 

the forms shown in the figure. 

237. Theorems of Taylor and Maclaurin : Infinite Forms. In 

Taylor's formula (B), Art. 229, we shall for brevity denote all the 
terms before R in the second member by P n , so that the formula 
may be written 

(B) f(a + h)=P n +R. 

P n is a polynomial of nth degree in h. Suppose we are able to 
determine the numerical values of f(x),f (x), . . . / (n) (x) for x = a. 
Then, for any assigned value of h, P n can be determined, and 
formula (B) would then enable us to determine the numerical 
value of f(x) for x = a + h, were it not for the fact that the value 
of R cannot be found because of the unknown 6. We may, how- 
ever, be able to show that R is very small as compared with P n , 
and in that event P n is itself an approximation to the value of 
/(V+ h). Such was the case in the example at the end of Art. 227, 
and in exercise 4, Art. 228. In fact, we there saw that by taking 
n large enough we could make R as small as we pleased, and would 
thus have in P n an approximation as close as we pleased to the 
true value of f(a + h). But R is not always small compared 
with P n . R may exceed, and often does exceed, P n in magnitude, 
and may even increase as n increases. In such a case the numer- 
ical value of /(a + h) cannot be determined by Taylor's formula, 
although that formula may still be useful in studying the properties 
of f(x), as it was found to be in Arts. 235 and 236. On the other 



§237 THEOREMS OF TAYLOR AND MACLAURIN 367 

hand, it may be that, under suitable choice of a and h, lim R = 0. 

71 = 00 

In that event, when n is taken large enough, P n will approximate as 
nearly as we please to the true value of f(a + h) . When lim R = 0, 

71=00 

we may omit R from formula (B), writing in the second member P n 
followed by . . . , or by the words "etc., etc." Thus 

f"(a\ f"'(d) 

(BO f{a + h) = /(a) + /'(a) h + J -^ h* + J ~^r h * + ■ • • 

n\ 

In like case Maclaurin's formula becomes 

f'(0) f'"(0) 

(CO f{x) = /(O) +/'(0) x + J -^p x' +Lp x> + . . . 

fln)(0) 






These are termed the infinite forms of the formulae of Taylor 
and Maclaurin to distinguish them from the finite forms (B) 
and(C). 

Observe that (B) is much more general than (BO ; for the hypoth- 
eses upon which (B) rests are that /Or) and its first n+1 derivatives 
shall be continuous throughout a prescribed interval, while the hy- 
potheses upon which (BO rests are that fix) and all its derivatives 
shall be continuous throughout this interval, and the additional 
hypothesis that lim R shall be 0. Similar remarks apply to (C) and 

71 = 00 

(CO- The conditions of continuit}^ are fulfilled by all the elemen- 
tary functions for nearly all values of their arguments, and therefore 
(B) and (C) hold for all the elementary functions, and for nearly 
all values of a and h (or x in case of (C)) . The condition lim R = 0, 

71 = 00 

however, is usually not fulfilled except for certain very special 
intervals, as we shall see in the examples that follow. Formulae 
(BO and (CO are therefore greatly restricted in their application. 
The difference between (B) and (BO (or between (C) and (CO) may 
be illustrated by the following example. From the identity 

1 == (l + h + h 2 + • • • + h n ) (1 - h) + h n+1 



368 CALCULUS §237 

follows, by division by 1 — h, 

08) r^r = 1 + h + A 2 + A 3 + . . . + h» + 



1 -A *■"''•■•• » « J _^i 

a formula which plainly holds for all values of h. This equation is 

fon+l 

analogous to Taylor's formula (B) , the fraction _ , correspond- 
ing to Taylor's remainder R. If we discard this fraction, and 
write "etc., etc.," or . . . after h n } we shall have a formula 
analogous to (B'), 

03') j-L-. = l + h + h? + h* + h* + . . . + h* + . . . . 
This equation asserts that if we take terms enough the second 
member is a close approximation to the true value of r- That 

this is true for some values of h we find by trial. Thus, on sub- 
stituting | for h in (/3'), we get 

2 = 1 +-2 + l + i + Ts + ■■■+¥+■■■' 

which tells us that the infinite geometric progression in the second 
member has the sum 2, and this we know, from elementary alge- 
bra, to be true. 

Again, when h = T \, (0') becomes 



1 
~9~ 



1 + TV + T*tf + tJuV + • • • = l.llH 



an equation which is obviously true. 

On the other hand, there are values of h, for which (/3') is not 
trujef. For example, when h = — 1, (/3') asserts that 
1=1-1 + 1-1 + 1-1 + 1 . . . , 

which is manifestly absurd. Again, when h = 2, (/?') becomes 

-1 = 1+2 + 4 + 8 + 16+..., 
which is also untrue. If, however, we substitute — 1 and 2 for h 
in (j3), we get 

i_l_l + l_l+... +(-l)*+ V X 2 ; , 
and -1 = 1 + 2 + 4 + 8 + . . . + 2» + 2 



1' 



§238 THEOREMS OF TAYLOR AND MACLAURIN 369 

results which are perfectly true for all values of n. The simple 
explanation of it all is this: When we divide 1 by 1 — h, and 
discard the remainder, the quotient will be approximately correct 
only in case the discarded remainder has the limit as we proceed 

fon+l 

farther and farther with the division. Now lim _ , = only 

when — 1 < h < 1. Hence ($') is valid only when —l<h<l. 
In the same way, while the finite forms of Taylor's and Maclaurin's 
formulae (B) and (C) hold for all values of a, h, and n (which satisfy 
the conditions of continuity), the infinite forms, (B') and (C')> are 
true only when lim R = 0, and this will usually be found to happen 

only for values of a and h within certain restricted intervals. 

238. Examples of the Application of the Infinite Forms of the 
Theorems of Taylor and Maclaurin. 

Example 1. The development of sin x. 
In Art. 230 we derived the development 

3 5 7 sm ( n ^) 

(SO ■n«-»-fj+f-f r H...+_L3Z,. + a, 



ft! 

n+1 



where R = sin [~(n + 1) £ + te~| • , *" '.. • 

L 2 J (n + 1) ! 

We shall show that lim i£ = 0, for all real values of x. 

n=oo 

Consider the factor - — ■ Let ft' and p be two positive integers such 

+ 1)1 

that n = n' + p. Then n + 1 = n' + 1 + p, and, whether x be + or — , 
we may write 

jz]«+i_ = UK+i _M_ _M is! 

(n+l)! (ft'+l)! ft'+2 ft'+3 ft'+l+p (? 

and therefore 

0SBm7 i^ SiBm? if£±i.fi£iy. 

n =oo(ft + l)! p =oo (ft' + l)! \n ' J 

Now choose n' >\x\ and hold ft' constant. Then lim ( ^-} ) =0, 

p=oo I w'/ 

whence lim ',', 1 ,, . f 1 ^ 1 = 0, and lim , ]x \ n = 0. Now the sole 

p = oo (ft'+l)! \ft'/ n =oo(w + l)I 



1 ZK+ 1 /|xj\ p 



370 CALCULUS §238 

difference between - — p-77 and - — ——. ■ is that the former is always + , 
(ti + D! (n+l)I ' 

while the latter, when a: is — , is alternately + and — as n increases. 

Therefore lim - — —— = for all real values of x. 
„=oo (ft + 1)1 

The other factor in R is sin (n + 1) ^ + 6x \; whose value always lies 
between —1 and +1. We conclude, then, that lim R = 0, for all real 

n=oo 

values of x. We are therefore warranted in writing 

/V»3 /V*5 /yi7 /y.9 

sin* =*__+-__+__ .... 

By means of this development a table of natural sines may be computed. 
x must of course be expressed in radians. 

Example 2. The development of log (1 + x). 
In Art. 230 we found that 

(I/) log(l + aO=z-| + |-f+ • • • + (-l)- 1 ^+#, 
where R = 



(n + 1) (1 + 0aO»+i 
We shall show that lim 2? = 0, when < x = 1. 

n=oo 

From < < 1, and < z = 1 follow 

0< r-r~ <1 and 0< , , , N ^ +1 — — - • < 



1 + 0:c (ft + 1) (1 + te)»+i w + 1 

Hence = lim 7 — , — - , —77 = lim — — = 0. /. lim R = 0. 

n =oo (71 + 1) (1 + 6x) n + l n =ooft+l n = oo 

It can also be shown that lim R = 0, when — 1 < x < 0, but the proof 

I n=oo 

is difficult and we omit it. Therefore the formula 

/v»2 n& /v»4 /v.5 

log(l + x)=^-- + i -- I + -- . . . 

is valid so long as — 1 < x = 1. For example, if we set x = 1 in this 

formula, we get 

i o 1 1.1 1.1 1 I 

^2 = 1-- + --^---+ ...., 

a formula by which the natural logarithm of 2 can be computed to any 
degree of accuracy by taking terms enough. However, as we saw in 
exercise 2, Art. 231, a very large number of terms are needed for great 
accuracy. On a later page we shall learn a more expeditious method of 
computing natural logarithms. 



§238 THEOREMS OF TAYLOR AND MACLAURIN 371 

Example 3. The Binomial Theorem. 

We employ Taylor's theorem (B) to develop (1 + h) k , where k is any 
real exponent. We have 

f( x ) = x k , f{x) = kx k ~\ fix) = k{k - l)x*-*, . . . 
f^ix) =k(k-l) . . . (& - n + l)s*- w , 
and f+Di x ) = k(k - 1) ... (k - n)x k - n ~K 

When k is — , f(x) and all its derivatives are go for x = 0. When k is + 
but not an integer, f n ^(x) is oo for x = if n > k. For all finite values of 
x save x = 0, fix) and all its derivatives are finite and continuous what- 
ever be the value of k. We have only to exclude x = from the interval 
to have the conditions of continuity fulfilled. Now the a of Taylor's 
formula (B) is 1 in this case, and we have 

/(i)=i,/ / (i)=^,r(i)=^-i),r , (i)=^-i) ik-2), . . . , 

/(n)(i) =kik-l) . . . ik-n + 1), 
and f+^il + dh) =kik-l) . . . (k - n)(l + dh) k - n ~K 
Substituting in (B), we have 

(1 + h )k = i + kh + k ( k ~ *) h t + M ~ 1) it ~ 2 ) h s + . . . 

^ ! o ! 

_|_ ^- 1) • • • (k-n+1) h n + R 

where g , d-'W-D ; ■ • (t-^' (1 + ^ l| 

ft! 

only the second form of the remainder being written because it is in this 
case the more useful. 

When k is a positive integer and n > k, f) ix) = 0, and the development 
terminates with the (ft + l)th term. We have then the ordinary binomial 
expansion. 

Suppose, then, that k is not a positive integer; then the development 
does not terminate, but there is a value for R for every value of ft. We 
shall show that lim R = for | h \ < 1. 

n=oo 

R may be written in the form 

R _ M-i)(t-2) ...(*-„) \(±^fr (1 + rt) « *. 

ft! L 1 i 9h J 

If in this we set z = . , / , R may be written 

B = *(*-l)*-(|-lW|-l)« • • - .^-lWl + 0/&)*"U. 



372 



CALCULUS 



238 



Because [ h | < 1, or — 1 < A < 1, it follows that 



Now set 



(1 - e)h 
1 + dh 



<1. 



P n = ^-l),.(|-l),....g-l), 
We shall show that lim P n = 0. 

71 = °0 

Let n = ft' + p, n' being a constant, and both ft' and p being positive 
integers. Then 

p„.-*(*-i)..(|-i)i — (J-i) 2 - 



n'+i 



p -'-=Mvt-i- i >-(vV2- i >' 

^^"•(vTI- 1 ) 2 -^- 1 )- • • • fcV 1 ) 2 - 

JFften & is +, we choose ft' such that n f + 1 >k, whereupon 

\(-T ] J-- 1 ) z \<\ z \ 
I Vft' + r / I 

for all positive integral values of r. Consequently 

0<|P n ' + p|<|Pn'|-|^| P . 

Because | z | < 1, lim \z\ p = 0, and therefore 

p = oo 

lim P n = lim P n '+ P = 0. 



TF%en & is 



> 1 and 



gral values of ft. But as ft = oo, 



( 1 )z > I z I for all positive inte- 

\ft I \ 

— 1 decreases towards the limit 1 . 

Hence, since \z\ <1, we can always find a value of ft, say ft', such that 

— — 1)2 shall be less than 1, although greater than \z\. Suppose that 
n I \ 

(-, — 1)2=5, where | z \ < 5 < 1 ; 

| Vft / | 

then ( -7—; 1 12 < 5 for all positive integral values of p. Therefore 

I \n + p I \ 

0<|P»'+pl<|iV|-|5| P . 



§238 THEOREMS OF TAYLOR AND MACLAURIN 373 

But lim | 5 \ p = 0, and consequently 



v- 



lim P„ = lim P n '+ P = 0. 

n—<x> p=oo 



We have now proved that lim P n = 0, for all values "of k. 

rt=00 

The next factor in R is (1 + dh) k ~ 1 , which varies with n because 6 does. 
Its value is, however, obviously finite. The remaining factor in R is the 
constant h. Therefore 



limfl = Y\mP n X\m{l + eh) k - l h = 0, 

n — oo n = oo n = oo 

-provided \h\ < 1. Therefore the formula 

(1) (1 + h) k = 1 + kh + k{k ~ l) K + k ( k -VJ k ~ 2 ) h i + 

^ A(t-l)...(fe-n+l) A 
w! 

is valid when — 1 < A < 1, and for all values of k. 

From (1) we get the binomial theorem for (a + b) k . When \a\ >\b\, 

we write {a + 6) fc = a k l 1 + - ) , and, since 
1 + -) by (!)• The expansion is 



< 1, we may expand 



K)'= '+'©+«^ j »er+ 



, A: (7b - 1) . . . (fc - n + 1) /6\" , 
n\ \al 

Multiplying both sides of this equation by a k , we get 

(2) \a\>\b\, 

(a + b) k = a k + ka k ~ l b + k(k . 1} a k ~ 2 6 2 + 

m + fc(fr - 1) . . . (£ - n + 1) a &-n 6 n + _ 
n! 

When |6| >|a|, we write (b + a) A = 6*( 1 + ? ) , expand the last 
factor by (1), multiply the result by 6*, and shall then have 

(3) |6|>|a|, 

(6 + a) k = 6* + *&*-*«! + k ^ k 1} &*" 2 a 2 + 

a! 

> ! fe(fe - 1) . . . (A; - n + 1) &fc _, jW j § 



374 CALCULUS §§239-240 

We have now proved the binomial theorem for any real exponent 
whatever. 

In applying the binomial theorem it must be borne in mind that, 
when the exponent k is not a positive integer, the expansion must 
proceed with descending powers of the greater of the two terms 
of the binomial, and with ascending powers of the lesser term. 

239. Exercises. 

1. Expand V I - x 2 ; Vl - e 2 cos 2 e. (See Art. 168, exercise 12.) 

2. Expand </l - x 2 for x 2 < 1 and for x 2 > 1. 

3. Show that when x 2 < 1, 

(l-x 2 p = 1+ ;U 2 + r^ 4 + £^-,x« + 3 ' 



Vf^ 2 2 2 -2 2 3 -3! 2 4 4! 

^ " " ' ^2 2W (ti!) 2 I" • • • • 
In the developments of the following functions, show that lim R = 0, for 

n=oo 

all values of x. 
4. cos #. 5. e x . 6. sin 2 z. 7. # sin #. 8. x cos a:. 

240. The Power Series. 

Definition. A never-ending succession of monomials in x, united 
by the signs + or — , 

(a) a + a\X + a 2 x 2 + a s x 3 + . . . + a n x n + . . . , 

where the constant coefficients, a i} proceed according to a known 
law, so that any coefficient can be calculated when its position in 
the series is known, is termed an infinite power series in x, or more 
briefly, a power series in x. 

A power series may be regarded as a polynomial in x of infinite 
degree. 

The second members of the following equations are power 
series : 

/v»3 /v»5 sy7 , / y2n — 1 

(1) *>,_._*+*_* + ...+ ( _ ir - 1{ ___ + . . . . 

(2) ..-1 + , +*+*+... +£,+ .... 

(3) io g (i + x) = x-| 2 +| 3 -^+ . . . +(-D n - 1 J+ 



§240 THEOREMS OF TAYLOR AND MACLAURIN 375 

The second member of Taylor's formula (B') (the infinite form) 
is a power series in A. It is characteristic of a power series that the 
coefficient in the nth term can be defined, explicitly or implicitly, 
as a function of n. 

Let P n be the sum of the first n terms of the power series (a), 

P n = a + dix + a 2 x 2 + a 3 rc 3 -{-...+ a n x n . 

If lim P n = A, where A is a finite and determinate constant, the 

n=°o 

series is said to be convergent or to converge to the limit A. A is 
also, though inaccurately, styled the "sum" of the series. If A 
is infinite or indeterminate, the series is termed divergent. A 
power series in z may be convergent for some values of x and 
divergent for other values of z. For example, the power series 

1 + z + z 2 + x 3 + ... + x n + . . . 
is divergent when x = 1, for then lim P n = go . It is also divergent 

n==oo 

when x = — 1, for then P n = 1 or 0, according as n is odd or even. 
On the other hand, the series is convergent when x = J, for we 
have then 

1+- + -4--4- — 4- • • • 4- — + • • • • 

and we recognize this as an " infinite geometrical progression " 
whose "sum" is 2. Indeed, from what was said in Art. 237, it 

appears that this power series converges to the limit ^ > when 

x X 

\x\ < 1, and diverges when | x | = 1. In this case we say that 
the interval of convergence is — 1 < x < 1. 

Let us again consider the finite and infinite forms of Taylor's 
formula, 

(B) f(a + h) =P n + R and (B') f(a + h) = P n + . . . . 

We have learned that (B r ) is valid when and only when lim R = 0. 

The examples of the foregoing pages showed, however, that the 
determination of the limit of R was often beset with difficulties. 
We shall now show that there is a way of determining the validity 
of the development (B r ) without making any use of R at all. The 



376 CALCULUS 

facts in the case are these : P n + . . . is a power series and is 
convergent when lim R = 0, and divergent when lim R ^ 0. 

n = °o n=oo 

For, (B) may be written in the form 

f(a + h) = limP n + lim£, 

n— oo n=°o 

and when the last limit is 0, we have 
(B) f(a + h)=l]mP n , 

71 = 00 

which asserts that when lim R = 0, P n converges to the limit 

71 = 00 

}{a + h). That P n is divergent when lim P ^ we shall accept 

71=00 

without proof.* We see then that (B) is only another way of writ- 
ing (B'), and that the second member of (B') has meant all along 
a power series in h. And, therefore, to say that (B') is valid pro- 
vided only that lim R = 0, is precisely equivalent to saying that 

n = °0 

(B') is valid or not, according as the power series P n + • • • is con- 
vergent or divergent. Now while the determination of the limit 
of R carries with it the determination of the convergence or diver- 
gence of P n , there are numerous tests for the convergence and 
divergence of P n which do not depend upon R, and when we can 
apply these tests we are able to avoid the difficulty of determining 
the limit of R. For example, in the binomial expansion (1), 
example 3, Art. 238, the second member can be shown, by the 
test for convergence presently to be explained, to be conver- 
gent for | h | < 1. Knowing this, we can pronounce formula (1) 
to be valid for such values of h without going through the difficult 
process of determining the limit of R. All the foregoing discussion 
applies, of course, to Maclaurin's formulse (C) and (C). 

* It is entirely conceivable that R should have a finite limit, not 0, and P n 
have at the same time a finite limit. This limit would not be f(a + h) but 
f(a + h) — lim R. It can be proved, however, that such a state of affairs 
cannot exist when f(x) is one of the elementary functions. For such func- 
tions, as stated in the text, P n is divergent and not equal to f(a + h) when 
lim R ^ 0. 



§241 



THEOREMS OF TAYLOR AND MACLAURIN 



377 



241. The Ratio Test for the Convergence of Power Series. 
A full treatment of the subject of convergence of power series 
would be out of place in an elementary textbook on the calculus,* 
and we shall therefore merely state, without proof, one of the 
most useful tests for convergence, viz., the ratio test. 

The ratio test is as follows: 

(n + l)th term 



Iflim 
Iflim 

71 = CO 

Iflim 



nth term 
(ft + l)th term 



nth term 



(n + l)th term 



< 1, the series is convergent. 

> 1, the series is divergent. 

= 1, no conclusion can be drawn: the 



nth term 

series may be convergent or divergent. 

Observe that, in order to establish the \ c o nver § ence f a . power 

\ divergence ) r 

i{ ! 1 f or 



(ft + l)th term 



series, it is not sufficient to show that 

nth term 

all values of n; it is the limit of the ratio that must be, in absolute 

;!■• 

Examples. 

1. In the series for sinz, example 1, Art. 238, the nth term is 



value, j 



( g„-Di • and the ( " + 1)th term is (drfryr 

Hence 

(n + l)th term = _ (2n- 1)! 2 = -x 2 

nth term (2n + l)!* ' 2n(2n+l)* 

— x 2 

And lim - — — — — = for all values of x. Therefore the series is 

n =oo 2n(2n + 1) 

convergent and the development valid for all values of x. Compare this 

proof with that given in example 1, Art. 238. 

2. In the series for (1 + h) k , (1), example 3, Art. 238, the nth and 

(n + l)th terms are respectively 



Kk-1) . . . (ft-n + 2) ^ and k{k - 1) 
(n- 1)! 



. (k - n + l) An 



* An excellent elementary presentation of the subject of series is Osgood's 
Introduction to Infinite Series, published by Harvard University. 



378 CALCULUS . §§242-243 

Then 

(n + l)th term _ k — n + 1 -, _ I _ -, , k -f- I V 
nth term n \ n / ' 

lim 



and 






z = ( 1 + - )£ and lim ( 1 + - )x 

n \ nj n =oo V n l 



Hence, by the ratio test, when | h \ < 1 the series is convergent and the 
expansion valid, when | h | > 1 the series is divergent and the develop- 
ment not valid, while when | In | = 1 we cannot say whether the series is 
convergent or not. Note the simplicity of this proof of the validity of 
the binomial expansion as compared with the proof given in Art. 238. 

The ratio test applies, of course, to any power series, whether it 
be derived by Taylor's theorem or not. 
3. In the power series, 

l + 2z + 3;r 2 + 4z 3 + . . . + nx n ~ l + 
(n + l)th term 
nth term 

Hence the series is convergent when | x \ < 1 and divergent when \x\ > 1 . 
The interval of convergence is — 1 < x < 1. 

In order to apply the ratio test, we must know the formula 
which expresses the nth term as a function of n. To obtain this 
formula is not always a simple matter. 

242. Exercises. Determine the interval of convergence of 
each of the following power series: 

1. l+z + 2!z 2 + 3!a; 3 + 4!;r 4 + . . . . 

/y»3 sy*0 <Y*l 

2.x + ^- + ^p + -^-+ .... 

) V3 V5 V7 

o i , 1 .1*3 , , 1-3-5 3 , L3.5.7 4 , 

3 - 1+ 2 X+ 2^ 2 :2 + 2T4T6* 3 + 2 TI7^8* 4 + ' ' • ' 

i_^_i.£ 4 _!l3 x 6 1-3-5 tf_ 
2 2*4 2-4*6 2- 4- 6*8 

Determine the intervals of convergence of the developments of the 
following functions found on the preceding pages: 
5. cos x. 6. log(l+z). 7. e x . 8. sin 2 a; 

243. Development of /(as) when a Formula for /( n >(#) Cannot 
Easily be Found. In such a case, although we may not be able 
to obtain a general formula for/( n )(z), we are often able to calcu- 



§244 THEOREMS OF TAYLOR AND MACLAURIN 379 

late numerical values of f(a), f(a), /"(a), . . . , and so to obtain 
as many terms of the development of /(a + h) as we please. Thus, 
let us obtain the development of tan x by Maclaurin's formula. 
The calculation of the first few derivatives of tan x will convince 
the student of the difficulty of expressing D n tan x in terms of 
trigonometric functions of x. For convenience we denote tan x 
and its derivatives by u, u', u", . . . . 
Then 

u = tana:, u' = sec 2 x = 1 + u 2 , \u" = uu' ', 
\ U '» =uu" + u' 2 ; 



u 1Y = uu'" + 3u'tt", 
hu y = uu IV +4cu'u"'+3u 



hi 



2 

\ u VI = uu y +5 u'vF + 10 u"u" , 
a u vn = ww vi _|_ 6 w ' w v _|_ 15 U 'V V + 10 w'" 2 , 
| w VIII = uu yn + 7 w'w VI + 21 u"w v + 35 u'"u IV , 
etc., etc., etc., 

Hence when x = 0, 

u =0, u" =0, ii IV = 0, w vr =0, w VIII = 0, w x =0, . . . 

u' = l, w'" = 2, u*=2*, wvir = 2 4 .17, u™ =23.31, u™ = 2*.691 

Substituting in Maclaurin's formula (C) we have 

, 2 3 24 24.17 7 28.31 29.691 n , 

t£ i nx = x+yX*+- l x?>+- 1 ^-xT + — gy-a; 9 + jjj s u + 

We make no attempt to determine the validity of this develop- 
ment because we can determine neither R nor the law of the series. 

244. Exercises. Develop the following functions by Mac- 
laurin's theorem without attempting to find R or the law of the 
series: 

1. log sec x to 5 terms. 4. e sin x to 6 terms. 

2. e x sin x to 8 terms. 5. e cosz to 5 terms. 

Hint, u = e x sin x, n lv = — 4 u. 6. log (1 + sin x) to 5 terms. 

3. sec z to 5 terms. 

Develop the following functions by Maclaurin's theorem and 
use the ratio test to determine the interval of convergence: 



380 CALCULUS § 245 

7. tan -1 xton terms. 

Hint. u = tan 1 x, v! = • 

1 + x 2 

Then by actual division 

u' = 1 — x 2 + x 4 — a; 6 + x 8 — . . . , 
w" = -2 x + 4 x* - 6 x 5 + 8 x 7 - . . . , 
^"=_2 + 3.4x 2 -5.6x 4 + 7.8x 6 - . . . , 
^ IV =2.3.4x-4.5.6:r 3 + 6.7.8z 5 - 
etc., etc., etc. 

8. Set x = 1 in the development of tan -1 x, and thus compute ir. 

9. sin -1 a; to n terms. 

Hint, u = sin -1 x, u'= (1— .t 2 )~*. Expand this by the binomial the- 
orem and proceed as in exercise 7. 

10. In the above development of sin -1 x, set x = \ and thus compute tt, 

245. Computation of Logarithms. The power series for 
log(l + x), example 2, Art. 238, is not well adapted to the compu- 
tation of logarithms because the terms are alternately + and — 
and do not decrease rapidly, because, in other words, the series 
converges very slowly, and, as a consequence, a large number of 
terms of the series would have to be used to secure a close approxi- 
mation to the value of log(l -\-x). A formula very well suited 
to the computation of logarithms is derived as follows: 

We have 

. , . X 2 X 3 X* X 5 

log(l+x) = x- J+J -^ + ^- 

Changing x to —xm this equation, there results 

/y»2 /y«3 .7*4 0"5 

1 * - v **/ %As %K/ tA/ 

. log(l-*) = -*-2-3-4-5- • • •• 

The interval of convergence for each series is —1 < x < 1. Sub- 
tracting the second equation from the first, we get 

, 1 + x ( X 3 X 5 X 7 ) 

Now set x = ^ — - in this equation and restrict m to positive 

2m + 1 



§245 THEOREMS OF TAYLOR AND MACLAURIN 381 

values. This insures x lying within the interval of convergence. 
Then 

. 1 -\-x , ra + 1 , , , ... , 

log _ = log = log(ra + 1) — logm, 

and we have 

c i i 

(m) log (ra + 1) = log m + 2 ] _ . 1 + 



2m + 1 ' 3(2m + l) 3 

+ 5(2ra+l) 5 + ' * ' )' 

a formula which enables us to compute log (m + 1) when log m is 
known. The power series within the brackets is seen to converge 
very rapidly, so that a comparatively small number of terms of 
this series will furnish a very close approximation to the value of 
log (ra + 1). Formula (m) is valid for all positive values of m. 
Making m = 1 in this formula, there results 

log 2 = 2 J ^ + g-^3 + ^p + y^ + g-^ + jj-^jj + . . . | 
= .693 147 180 559 945. 
Making m = 2 in (m), we have 

^3 = ^2 + 2^ + 3^ + ^ + ^ + ^+... j. 

Or, making m = \ in (m) and noting that log (J + 1) — log \ = 
log 3, we have 

l0g3 = 2 H + ^ + 5T2^ + 772~ 7 + '• "j 
log 3 = 1.098 612 288 668109. 

The student should make all the computations for log 2 and 
log 3. These are, of course, natural or Napierian logarithms, 
the base being e = 2.718 2818. . . . 

Logarithms to any base, a, can be derived from the natural 
logarithms by means of the formula 

, A7 log e N 
log a N= f^—> 
log e a 



382 CALCULUS §§246-247 

where N is any member. To prove this formula, we take the 

natural logarithms of both sides of the identity 

N= a l °z« N 
and get 

log e N = log e a}°z° * = log a N • log e a, 

and the given formula follows by division. In particular the Briggs 

or common logarithms are given by the formula 

log A 7 " 



logioN 



log 10 



The value of 1 — -^ is .4342 9448. This number is termed the 
log 10 

modulus of the common system of logarithms. 

246. Exercises. 

1. From the natural logarithm of 2 and 3 of the preceding article com- 
pute the common logarithms of 2 and 3. 

2. Compute the natural logarithm of 10. 

Ans. log 10 = 2.302 585 092 994 045. 

3. Compute the natural and the common logarithms of 5, 7, 11, 13, 
17, 19, correct to 8 places of decimals. 

247. Theorems of Taylor and Maclaurin for Functions of Any 
Number of Variables. We begin with a function of two variables. 
Let /Or, y) and its partial derivatives of orders up to and including 
n + 1 be continuous within a region about {x , y ). We seek to 
develop first /(x + h, y + k) in powers of h and k. 

Let (j>(t) = f(x + ht, 2/0 + kt) = f(u, v), 

where u = x + ht, v = y + kt. 

Developing 0(0 in powers of t by Maclaurin's theorem, we have 

*(0-*(O)+«*'(O)+J*"(O)+ . . . + ^0W(O)+ ^Jjj0OH-i>(«). 

We now set t = 1 in this equation, and there results 
f(xo + h,yo + k)=4>(l) = 

0(O)+^(O)+^0 ,, (O)+ • . . +^ (n) (0) + (^ 1 y ! ^ +1 )W. 

Now, since <f>(t) = f(u, v) and u and v are functions of t,~ 
,,, A df(u, v) du df(u, v) dv df(u, v) df(u, v) 

{t)= —dvr Tt + — &r dt ~ h ~&ur + /c_ ^r * 



§247 THEOREMS OF TAYLOR AND MACLAURIN 383 

This is abbreviated to 

Also, since 4>'(t) is a function of u and v } we have 

4> W __ 0W _/ i __ + fc___ . 



But 



gj/(0 = g ( k df(u,v) k df(u,v) \ 
du du\ du dv J 

. _ h d*f(u,v) J: dy(u,v) 

du 2 du dv 

a - -i i W (t) 7 d 2 f(u,v) • d 2 f(u,v) 

Similarly, -^ = * -^ + * ->^±. 

Therefore 0"(*) = V *^ + 2 Wfe *J^ + * 2 ^J^ 

dw 2 dudv dv 2 



^ 2 + 2 ^^ + ^)/(v). 



^ . Q „ d d h2 d 2 
dw 2 dw dv dv 

The expression within the brackets looks like the square of 

A A \ 

& — + /c — j and we shall write it as such, understanding that 
-r— 2> ^-g) -^- ^-> are not real powers and product. We have then 



♦ w »-(*.i+*i) , ^ , >- 



In like manner it can be shown that 



(Let the student work this out.) 
This also is written in the symbolic form 

+"'®=( h Tu + k Tp^- 

Also in the same way it may be shown that 

and *<-)(<)= (A ^ + *|;)V(m,»). 



384 CALCULUS § 247 

Now, when t = 0, we have u = x 0} v = y , 0(0) = f(x , y Q ) and 

which means 

( h df(x, y) + k df{x, y)\ . 
\ dx dy ) x=Xo 

y=Vo 

Similarly, 
And finally, 



£=x +M 
2/=Z/o-HA; 



Substituting in the equation for f(xo + h, yo + &), we have 

'i^ + fc^— J/(^o,2/o) 

+ K* A + * wJ f{Xo ' yo)+ --- + ^{ h i +k wJ f{Xo ' yo) 

This is Taylor's theorem for a function of two variables. Mac- 
laurin's theorem for the development of f(h, k) in powers of h and 
k is obtained by substituting for x and for y in (B) . 

These theorems are readily extended to functions of any number 
of variables. Let f(x 1} x 2 , . . . x k ) be a function of k variables. 
Wei shall write Taylor's theorem for the development of 

f(xi + h h X2 + h 2 , . . . x k ' + h k ), 

where Xi, x 2 ', . . . x k , are the initial values of the z's, corre- 
sponding to the x , y of (B) . For brevity we write the functional 
symbols in the forms f(x) , f(x') , f(x' + h) . The bracketed symbol, 
which is in this case 

(*»S? + * i ai7 + •••+** as?)' - ■; 

we shall write in the compact form (h-^-j\ 



§247 THEOREMS OF TAYLOR AND MACLAURIN 385 

Then Taylor's theorem is 

(B') f(x'+h) =m + (h-^f(x') + ±(h-^)ff(x') 

+ 1 vK) 3f(x ' )+ ■ ■ ■ + ii{ h w) fix,) 



(n+i;i|_\ vu,/ /jx=x'+eh 

Example. f(x, y) = sinxy. to develop 

f(x .+ h, y -\-k) = sin (x +h) (y Q + &). 
Solution. 

ax 



ycosxy,— = xcosxy, whence (/& \~k — )= (hy -\-kx ) cos x y . 

dy \ dx dyoj 

d 2 f . . d 2 f 2 . a 2 / 

ax 2 y dy 2 v ' dxdy 



d 2 f - ■ a 2 / » ■ a 2 / , 

— iy2 sin r-?/ — *- = — x 2 sm a;?/, — — = —xy sm x?/ + cos x?/. 



(fc — - + fc — Y/fo 2/) = h 2 -^ + 2 hk 

\ dx dy j J ,y dxo 2 



Then 

dx d^o dy 2 

= - (/i 2 t/o 2 + 2 hkx Q y + ft 2 x 2 ) sin x ?/o + 2M cos x ?/o 
= — (hy + &x ) 2 sin x y -\-2hk cos x ?/o. 
Again, 

3 3 / 3 3 3 / 3 

3 — —y cos #2/ ? 3 = ~~ x cos x v> 

dx dy 

dH dH 

— f — = —xy 2 cos xy — 2 y sm xy, — J —- = —x 2 y cos xy — 2 x sm xy. 
dx 2 3?/ dx dy z 

Then 

/(x, ?/) = ¥ -'', + 3 h 2 k -^- + 3 M 2 -^- -h A; 3 • 
JV ,yj dx* dx 2 dy dxdy 2 dy 3 



( h A +k ±)% }y) = ¥ ^, sm ^ + shk ,^ +k ^ 

\ dx dy) JK ,yj dx 3 dx 2 dy dxdy 2 di 

= — (% + kx) z cos xy — 6 /i&(% + &x) sin xy, 



whence 



[Mi + »i)M, 



—a 3 cos (x + 0/0 (2/0 + 0&) — 6 /*&« sin (x + Oh) (y + 0&), 

where a = fa/o + ftx + 2 0M;. 

Substituting the foregoing results in (B), we have 

(b) sin (x + h) (y + k) = 

sin x Q y + (%o + kx Q ) cos x ?/o 4- ^ cos x y — § (%o + kx ) 2 sin x 2/o 
— I a [6 M sin (x + 0/1) (2/0 + 6k) + a 2 cos (x + 0^) (2/0 + Ok)], 

and this is the development sought. 



386 CALCULUS §§248-249 

Let us in this development set x = y = 0. There results 

(c) sin hk = hk- § dhk [6 hk sin 2 M + 4 0TO 2 cos 2 M] 

= hk-2 dh 2 k 2 sin 2 M - 1 3 M; 3 cos 2 /ift. 

Now set k = 1 in (c) and get 

(d) sin h = h - 2 0& 2 sin 2 /i - f 3 ft 3 cos 2 A. 
Again in (b) let us set h = 0, y = 0; there results 

sin #o& = kx — | fcW cos dkx . 

In this let A; = 1, 

/ x . cos0£ o q 

(e) sin z = Zo — — r: — z . 

o! 

Compare (d) and (e) with (S 3 ') of Art. 227. 

248. Exercises. 

1. Develop e(*o+ft)teo+A) f or n = 4. 

2. Set h = y = 0, and & = 1 in the development in exercise 1. 

3. Develop e x *+ h sin (y -\- k) for n = 4. 

4. Set /*. = ?/ = in the development in exercise 3. 

5. By formula (B') write out in full the development of f{x, y, z) for 
n — 0, 1, and 2. 

6. f(x lf x 2 , x z ) = anXj 2 + a 22 x 2 2 + a 33 x 3 2 + 2 a^-r^ + 2 a 23 x 2 :r3+ 2 a^Zs^, 
where a lfc = a kl . 

By formula (B') develop /(a;/ + h u x 2 + ^2, &»' + ^3). 

249. Euler's * Theorem for Homogeneous Functions. A ho- 
mogeneous function has been denned (Art. 186, exercise 16) as a 
function in which all the terms are of the same degree in all the 
variables together. This definition applies to algebraic functions 
only, A more general definition, applicable to any function, is as 
follows : 

A function is homogeneous if, when each variable in it is multi- 
plied by the same number, the function is multiplied by a power of 
that number. 

Thus, f(x, y) is homogeneous if 

f(tx,ty) =t»f(x,y). 
* Pronounced Oi'ler. 



§249 THEOREMS OF TAYLOR AND MACLAURIN 387 

The exponent n is termed the order of the function. For example, 

z — Vx 2 + y 2 , \— f- , ; , x 2 sin ir - > 

ex + dy x- Vz 2 + xy 3 y 

log xyz — log (x 3 + 2/ 3 + £ 3 ) » 

are homogeneous and of order 1, 0, 0, 2, 0, respectively. The 
function in exercise 6, Art. 248, is homogeneous and of order 2. 
Euler's Theorem. If f(x, y) is homogeneous and of order n, then 

Proof. Let <f>(t) = fitx, ty) = f(u, v), where u = tx,v = ty. Then 
by formula I of Art. 203, 

d<f>(t) _ df(u, v) du df(u, v) dv _ df(u, v) df(u, v) 
~~dT du di^ dv di ~ X ~du *~ y dv 

But, because f(x, y) is homogeneous, 

4>{t) = t»f(x, y) and ^ = nt^f(x, y). 



Therefore 



x ejM + y ejM = nt ^ f(Xyy) _ 



Setting t = 1, we have u = x, v = y, and 

(E) x ^y) +y ^) = ' nf(Xty) . q. e . d . 

For sake of brevity Euler's formula is often written in the form 

The theorem and the proof hold for homogeneous functions of 
any number of variables. For example, if f(x h x 2 , . . . x k ) is 
homogeneous and of order n in the k variables, Xi, x 2 , . . . x kj 
then 

(E ' } ( Xi i i +x >k + X3 -k+ • ■ ■ +**a9 /(Xl > • ■ • **> 

= »/(*!, . . . X k ). 



388 CALCULUS §§ 250-251 

250. Exercises. 

1. Verify the formula for each of the examples at the beginning of the 
preceding article. 

2. Carry out the proof of Euler's theorem for a homogeneous function 
in three variables. 

251. The Imaginary Exponent : De Moivre's Theorem. It has 

been shown that the following developments are convergent for 
all values of x : 



C0SX=1 -2i + Ii- 



X 2 X s X* 

M + 3! + 4! 

X* X 6 . X s 



= 1 + * + 2!+3! + Ii + 



smz 



2! '4! 6! ' 8! 

X 3 X 5 X 7 X 9 

= X ~ 3! + 5! " 71 + 9! 



In the first of these let id be written in place of 6, (i = V — 1). 
The result is 

Neither side of this equation has as yet any meaning, because no 
meaning has been assigned to an imaginary exponent, and because, 
further, we cannot yet assert that the series with imaginary terms 
in the second member has any meaning. But this equation can 
be brought to the form 

e 2 e* \ ./„ e 3 # 5 \ 

in which the second member is separated into two series that are 
convergent for all values of 6 and are developments of cos 6 and 
sin 6 respectively. This equation can therefore be written 

(A) e 10 = cos 6 + i sin 6, 

and the second member is seen to have a definite meaning. 
We now take this equation (A) as the definition of e w . 
Writing — i for i in (A) we have 

(A') e-* = cos 6 — i sin 6. 



251 



DE MOIVRE'S THEOREM 



389 



Since this relation can also be gotten by changing the 6 of (A) to — 0, 
(A') is no new formula but is included in (A). By giving in (A) 

the values 2 x, ir, „ , — ~ , -~- in turn, we obtain the following in- 
teresting relations: 

e 2*i — C os 2 7r + i sin 2 t = 1, 
e™ = cos w -\- i sin 7r = — 1, 



(B) 



e 2 = cos^ + ism= = *, 



e 2 = cos _ i sm _ __ ^ 

?? 3tt , . . Sir 

e 2 = cos -7T- + t sm -=- = — i 



The reader doubtless knows that every complex number can be 
thrown into the form p(cos + i sin 0), where p and are real. 
Thus 

a + H=Vtf + b*( y a +j b V 

\Va 2 + b 2 Va 2 + b 2 / 
a , b 



Now 



Va 2 + ¥ 



and 



Va 2 + 6 2 



are the cosine and sine of some 



real angle which we will call 0. Moreover, v a 2 + 6 2 is also real, 
and we will denote the positive value of this real square root by 

p. Then 

a + bi = p(cos 8 -\- i sin 0). 

The quantity within the brackets may now be replaced by its 
value as defined by (A), and there results 

a + bi = pe#. 

This tells us that every complex number can be expressed as a 

product of a real positive number p by an " imaginary power" 

of e* 

Let n$ be written for in (A), n being any real number. The 

result is 

e ine = C os nd -\- i sin nd. 



* p is termed the modulus and 6 the argument of the complex number. 



390 CALCULUS § 251 

But e ind = (e i9 ) n * = (cos + i sin0) n . Therefore 

(I) (cos 6 -\-i sin0) n = cosn0 + i sinn0. 

This important formula is known as De Moivre's theorem. It is 
interesting to see how De Moivre's theorem may be used to 
derive expressions for sin nd and cos nd in terms of sin and cos 0. 
For example, 

cos 3 + i sin 3 = (cos + i sin 0) 3 

= cos 3 + 3 i cos 2 — 3 cos sin 2 — i sin 3 0. 

Equating the real parts on the two sides of the equation, and like- 
wise the imaginary parts, we have 

cos 3 = cos 3 — 3 cos sin 2 = 4 cos 3 3 cos0, 
sin 3 = 3 cos 2 sin - sin 3 = 3 sin - 4 sin 3 0. 

And this method is effective for cos nd and sin nd for any positive 
integral value of n. 

Writing - for n in (I), we have 

- ft ft 

(V) (cos -\-i sin 6) n = cos - + i sin - . 

\ / v / n n 

This, of course, is not a new relation, but is included in (I). Now 
we have the identities 

cos ± i sin = cos (0 + 2 kir) db i sin (0 + 2 hr), 

i i 

and (cos ± i sin 0)» = [cos (0 + 2 A;tt) db i sin (0 + 2 /ctt)]", 

where k is any integer. Applying (F) to the second member of 

the last identity, we get 

n • ' AN" /0 2/C7T\ . . /0 2/C7T\ 

(II) (cos =b i sin 0) n = cos - H J =b i sin [ - H , 

v y \n n J \n n J 

k = 0, ±1, ±2, . . . . 

i 
When n is an integer, n different values of (cos0 + tsin 0) n 

can be gotten by giving ft in (II) the values 0, 1, 2, ... n — 1. 

i 
Any other value of k causes (cos0 + isind) 71 to take one of the 

* This means that we impose upon the imaginary number &# the condi- 
tion that it shall obey the laws of exponents established for real numbers. 



§251 DE MOIVRE'S THEOREM 391 

l 
values already found. Hence (cos0 + isin0) u has n different 
values and no more. 

Formula (II) enables us to calculate the n nth roots of any num- 
ber. 

Example 1. To find the three cube roots of 1. 

Since cos + i sin = 1, we set = and n = 3 in (II) and get 









<Ti- 


2kir . 

= cos— + 


. . 2kir 

ism — . 


When k = 


-o, 


<T\ = 


cosO+ 


isinO = 1. 






When k = 


--1, 


,</! = 


2tt 

cos- 


, . • 2 7T 

+ % sin — = 


2 


Vs 


WV»pn h = 


= 9 


a 3 /T = 


4tt 
p.ns 


... 4* 

4- i. sin = 


-1-1 


iVz 






For further values of k this series of values of \/l repeats. Hence there 

are but three cube roots of 1, and they are 1, . To find the 

n nth roots of any real number, we have only to multiply the real positive 
nth root of the absolute value (value without regard to sign) of that num- 
ber by the n nth roots of +1 or — 1. For example, the three cube roots 
of 7 are found by multiplying the ordinary arithmetical value of \^7 by the 
three cube roots of 1. 

Example 2. To find the two values of s/ —i. 

By formula (B), — i = cos ~ — i sin ^. Then by (II) 



■i= COS 



f 2~\~ kir) — i sinf^-h kir). 



When k = 0, V— i — cos -. — i sin 7 = - 



4 4 V2 

TT71. 7 1 .. / • 5tT .. 5tT — 1 + % 

When k = 1, V — i = cos — — i sin — - = — — . 

4 4 V2 

And for further values of k these two values of V— i repeat. 

Formula (II) can be used to find the n nth roots of any complex number. 

For we have 

a + hi = p(cos + i sin 0), 

and 

1 1 1 

(a + hi) ~ n = p" (cos d + i sin 0)\ 



392 CALCULUS §§252-253 

Then the n values of (a -+- bi) n may be gotten by multiplying the ordinary 

1 l 

arithmetical value of p n by the n values of (cos + i sin 0) n as determined 
by (II) 

252. Exercises. 

1. By De Moivre's theorem express sin40, cos40, sin50, cos50, in 
terms of sin and cos 0. 

2. Find the three values of v^i, and of V 7 — i. 

3. Find the four values of V 7 — 1. 

4. Find the five values of ^—32. 

5. Find the six values of V^. 

6. Find the two values of Vs + 4 i. 

253. The Hyperbolic Functions. Let ix be written for 6 in 
formulae (A) and (A') of Art. 251. The results are 

e -x = cos ix -j- i sin ix, 

e x = cos ix — i sin ix. 

From these we get 

. A . ... e x — e~ x . e x + e~ x . , . e x — e~ x 

(A) — ism 23 = _ , cosi# = ^ , — ^tan^£ = 

v J 2 2 e x + e~ x 

Just as (A) of Art. 251 defined the imaginary exponent,, so formulae 
(A) of the present article may be regarded as the definitions of the 
sine, cosine, and tangent of an imaginary angle. It is worth not- 
ing that cosix is real, while simx and tan ire are pure imaginaries. 
The second members of formulae (A) play an important part in 
some branches of mathematics. They are called hyperbolic func- 
tionsJ)f x, and are denoted by sinhz, coshx, tanhz. We have, then, 
as the definitions of the hyperbolic functions, 

c x — e~ x 
Hyperbolic sine of x = sinh x = — » = — i sin ix. 

Hyperbolic cosine of x = cosh x = ^ — = cos ix. 

gX g — X 

Hyperbolic tangent of x = tanh x = - . _ x = —i tan ix. 

The simpler notation, shx, chx, thx, may be used for these func- 
tions. 



§254 THE HYPERBOLIC FUNCTIONS 393 

The hyperbolic secant, cosecant, and cotangent are the recipro- 
cals of the hyperbolic cosine, sine, and tangent. That is, 

1 2 

Hyperbolic secant of x = sech x = — r— = sec ix = — ; 

J * cosh x e x + e~ x 

1 2 

Hyperbolic cosecant of x = csch x = . , = i esc ix = 






sinh £ e* — e~ x 

qx _L_ g— x 

Hyperbolic cotangent of x = coth x = t — r— = ^ cot ix = x _ _ x - 

Observe that all the hyperbolic functions of x are real when x is 
real. 

Problem 1. Verify the following relations: 

ch 2 x — sh 2 x = 1 ; sech 2 z + tanh 2 x = 1 ; coth 2 x — csch 2 z = 1. 

Problem 2. Prove the following developments: they could be taken as 
definitions of sh x and ch x. 

Bhx = x +'fi+$+ ■ ■■' 

~2 ~4 

cha; = l+ — ++.... 
2! 4! 

Problem 3. Prove the following: 

— sha; = chz: — -chx = shx; — thz = sech 2 z; 
ax ax ax 

—sech x = — sech z tanh z ; — cschx = — csch x coth z ; — coth x = — csch 2 z. 
Gte ax ax 

Compare these with the derivatives of the trigonometric functions. 
Problem 4. Prove the following relations : 

sh (x ± y) = sh x ch ?/ ± ch x sh y ; 
ch (x ±y)= ch x ch ?/ ± sh x sh y; 
th x =t th w 



th (x ± 2/) 



1 =fc th x th 



254. The Inverse Hyperbolic Functions. The number whose 
hyperbolic sine is x is called the antihyperbolic sine of x, or the 
inverse hyperbolic sine of x, and is denoted by sh -1 x. If y is this 
number, then y = sh -1 x, a statement exactly equivalent to x = sh y. 



394 CALCULUS §254 

In a similar way are defined ch -1 x, thr l x, and the other antif unc- 
tions. We have 

& — erv 
x = shy = ^ ' 

whence e 2v — 2 xey — 1 = and ey = x rb vT+7 2 . 

Since e^ is + for all real values of y, the sign + must be used with 

this radical, and hence 

ey = x + Vl + x 2 . 

From this we get 

y = sh" 1 x = log (x + Vl +x 2 ). 

Similarly, it can be shown that 



ch" 1 ^ = log (x =b Vx 2 - l), 
where both signs may be used with the radical. From the identity 



(x + Vx 2 - l) - Vx 2 - 1) = 1, 
follows log (x - Vx 2 -l)= -log (z + Vx 2 - 1). 

Therefore we may write 



ch _1 x = zb log (a: + Vz 2 — l). 

Problem 1. Show that th _1 x = - log 

2 1 — x 

Problem 2. Prove that 

/sh -1 x = ; — ch -1 # = ± — : — th -1 x 



dx Vl + x 2 dx Vx 2 — l dx 1 — x 2 

The following formula? express the antihyperbolic functions as 
definite integrals: 

■ log (x + Vl + x 2 ) = sh -1 x; 



f 

J i Vl + x* 

f X dx = log (x + Vx 2 - l) = ± ch- 1 z; 
Ji Vx 2 - 1 

J" z dx 1 , 1 + z ,, , 



§255 THE HYPERBOLIC FUNCTIONS 395 

Compare these with the trigonometric integrals 

C x dx . . . p dx , , „ 

/ — = sin -1 a; = —cos -1 x; I - — . — ^ = tan _1 x.* 

Jo Vl — x 2 J* 1 + x 2 

255. Geometrical Properties of the Hyperbolic Functions. 

The hyperbolic functions and their inverse functions are related 
to the equilateral hyperbola, x 2 — y 2 = 1, somewhat as the 
trigonometric and circular functions are related to the circle 
x 2 + y 2 = 1. To bring out the analogy more fully, we shall con- 
sider first the trigonometric functions. Suppose that sin and 
cos were entirely unknown to us except as they are defined by 
the power series at the beginning of Art. 251 : we should probably 
not represent as an angle, nor observe the relations of sin 6 and 
cos to the circle. The equation cos 2 + sin 2 = 1 could, how- 
ever, be derived from the power series, and from this relation and 
from the equation of the circle, x 2 + y 2 = a 2 , we could derive by 
inspection the equations 

x = a cos 8, y = a sin 6, 

which are parametric equations of the circle. The geometric 
meaning of the parameter 6 would remain to be ascertained. 

Now in just this way we may connect the hyperbolic func- 
tions with the equilateral hyperbola. Being given the relation 
ch 2 — sh 2 0=1, and the equation of the curve, x 2 — y 2 = a 2 , we 
can write down at once the equations 

x = a ch 6, y = a sh 0, 

and in these we have a pair of parametric equations of the curve. 
The geometric meaning of the parameter will be determined 
presently. 

Consider next the unit circle, x 2 + y 2 = 1, and let u be the 
area of the shaded sector, whose arc is 0. Then 

u = | 0, = 2 u, 

* In Peirce's Short Table of Integrals may be found a collection of for- 
mulae relating to hyperbolic functions. 



396 

and 



CALCULUS 



x = cos 6 = cos 2 u, y = sin 6 = sin 2 w, - = tan 2 w, 



255 



P(x,y) 




and here we have x, y, and - ex- 

x 

pressed as trigonometric functions 

of the double area 2 u. 

Following the method presently 

to be employed with the hyperbolic 

functions, we shall derive these very 

simple relations again by means of 

definite integrals. 



We have 



u = area OAPB — ^ xy. 
Area OAPB = H "x dy = / Vl - y 2 dy 

Jy=0 JO 

= - (yVl - y 2 + sin" 1 ^) = -xy + ^sin-^. 



Hence 



1 . 



sin -1 y and y = sm2u. 



Further, 



= VI 



x = v i — y 2 = Vl — sin 2 2 u = cos 2 u, 

and we have the same expressions for x and y as before, but we 
have obtained them here without making any use of the repre- 
sentation of 6 as an angle. 

Consider now the equilateral hyperbola x 2 — y 2 = 1, and let u 
be the area of the shaded sector. Then 

u = area OAPB — ~xy. 
Area OAPB = f^x dy = f Vl + y 2 dy 

Jy=0 JO 

= \ [2/Vl + y 2 + log (y + VF+y 2 )] 
= ^ xy + ^ log (y + vT+l/ 2 "). 



§255 
Therefore 



THE HYPERBOLIC FUNCTIONS 



u = g log (y + Vl + i/ 2 ) = - sh" 1 y, 



397 



P(»,y) 




and 

Further, 

We have then 



y = sh 2 m. 

a; = VT+Y 2 = Vl + sh 2 2 w = ch 2 w. 

^ = ch 2 w, ?/ = sh 2 1£, - = th 2 w, 



which express the # and ?/ of a point on the equilateral hyperbola 
as hyperbolic functions of the double area 2 u. We see then 
that 6 is twice the area u. 

Problem. Show, by the same method, that when the equation of the 
curve is x 2 — y 2 = a 2 , 



2u 



2u 



x = a ch^ , y = ash— - 



CHAPTER XXXII 
INTEGRATION OF RATIONAL FRACTIONS 

256. Properties of Polynomials. We shall state a few of the 
properties of polynomials that we shall have occasion to use, re- 
ferring the reader to algebra for the proofs of these properties. 

Let 

(1) P(x) = a x n + ciix 71 - 1 -\- a 2 x n - 2 + . . . + a n - 2 x 2 + a n - 1 x + a n 

be a polynomial of the nth degree in x, the coefficients, a , a h . . . 
a n being real constants. Such a polynomial has n roots. Let 
fi, r 2 , . . . r n be these roots. Then it is a principle of algebra 
that 

P(x) can be resolved into a product of n linear factors of the form 
x — r; that is, 

(2) P(x) = a (x - ri) (x — r 2 ) . . . (x - r n ). 

The n roots may be all different, or there may be equal roots 
among them. A form that brings into evidence the repeated 
roots is 

(2') P(x) = a (x — ri) s i(# — r 2 ) S2 . . . (x — r k )\ 

where the s's are all integers and S\ + s 2 + . . . s k = n. 

The roots, r, may be all real, or some of them may be imaginary. 
Suppose that a + ib is one of the imaginary roots. Then it is 
proved in algebra that the conjugate imaginary a — ib is also a 
root. That is, 

The imaginary roots of a polynomial occur in pairs. 

Hence two of the factors of P(x) are x — a — ib and x — a + ib. 
The product of these factors is 

(x — a) 2 — (ib) 2 = x 2 — 2 ax + a 2 + b 2 = x 2 + px + q, 

398 






§257 INTEGRATION OF RATIONAL FRACTIONS 399 

a quadratic whose constants, p and q, are real. From this follows 
the important property that 

P(x) can be resolved into a product of real factors, linear and quad- 
ratic; that is, 

(3) P{x) = a Q (x - n) (x - r 2 ) . . . (x - r k ) X 

(x 2 + pix + gi) (x 2 + p 2 x + q 2 ) . . . (x 2 + p m x + q m ), 

where the r's, p's, and g's are all real, and k + 2 m = n. A form 
which brings into evidence the repeated roots is 

(3') P{x) = a (x - r^ (x - r 2 ) s * . . . (x - r k ) s > X 

{x 2 +piX J rqi) t i(x 2 + p 2 x-\-q2) t * . . . (x 2 + p m a; + gm)' m , 

where the s's and t's are integers and 

si + s 2 + ... + s& + 2 (*i + i 2 + . . . + t m ) = n. 

Now a rational fraction is one whose numerator and denomi- 
nator are polynomials, and therefore, by the principles just 
stated, the denominator of a rational fraction can always be 
resolved into a product of linear factors real and complex, or into 
a product of real factors linear and quadratic. Of these properties 
we shall make immediate use. To actually resolve a given 
polynomial into its factors is a problem that belongs to algebra. 
The student will not be called upon here to solve this problem 
except in a few simple cases, because in the exercises that follow 
the factors of the denominators of the rational fractions are either 
given or are easily determined. 

257. Integration of a Rational Fraction when the Denomi- 
nator is Resolved into Its Linear Factors. A rational fraction 
can be resolved into an algebraic sum of simpler fractions whose 
numerators are constants and whose denominators are the linear 
factors of the denominator of the given fraction. For example, 

2 x 2 + 9 x — 27 

the fraction ^ — -x can be shown to be equivalent to 

x y x 

3 1 2 

- -\ — ^ . Each of these last fractions can be inte- 

x x — 6 x -\- 6 

grated, and so the integral of the given fraction can be obtained. 
In integrating a rational fraction, the first step consists, then, in 



400 CALCULUS §257 

separating the fraction into its component or so-called "partial" 
fractions. The method of doing this can be learned by studying 
the illustrative examples under the several cases that follow. 

Case 1. None of the linear factors of the denominator are repeated. 

Example 1. To evaluate 

2s»-s + 3 _ d 
Jxix 2 - 1) (2s -3) ' 

The linear factors of the denominator are z, x — 1, x -f- 1, 2 x — 3. 

We now assume the identity 

(1) 2x 2 -x-\-S A, B C D 

{) x(x 2 -l)(2x-3) x~*~ x-l~*~x+l 2x-Z' 

Clearing of fractions 

(2) 2 x 2 - x + 3 s A(x 2 - 1) (2 x - 3) + Bx{x + 1) (2 x - 3) 

-hCz(x - 1) (2 x - 3) + £z (z 2 - 1). 

Since by assumption, (1) and (2) are identities, they hold for all values of 
x. Now by giving to x successively any four values we shall get four 
equations of first degree in A, B, C, and D, from which the values of these 
quantities can be determined. We choose such values for x as will 
render the resulting equations in A, B, C, D as simple as possible. 

When x = 0, 3 = 3^1 whence A = 1. 

When x = 1, 2 - 1 + 3 = 4 = -2 B whence B = -2. 

When x=-l, 2 + l + 3 = 6 = -10C whence C = - % • 

5 

When x = |, ~- | + 3 = 6 = ^D whenceD = ~ 

Therefore 






3 16 



(os 2x 2 -x + 3 = 1 _ _2 5_ - 5 

{ } x(x 2 - 1) (2x - 3) x x- 1 s-fl2s-3* 

We find by actual calculation that the algebraic sum of the fractions in 
the second member of (3) is equal to the first member, and therefore our 
original assumption is justified, and the given fraction is actually separated 
into partial fractions. Integrating (3), we have 

/ 2 x 2 — x + 3 , _ rdx _ 2 r dx 3 r dx , 16 r dx 
x(x 2 -l)(2x-S) X ~J x Jx-1 5 J.T + r 5 J2x-3 

= logs-21og(x-l)-|log(s+l) + |log(2a;-3). 



§257 INTEGRATION OF RATIONAL FRACTIONS 401 

Therefore 

r 2*-* + 3 dx = l *(2s-3)l 

J x(a; 2 - 1) (2 a; - 3) (a; - l) 2 (x + 1)* 

The methods of this and the following cases apply only when the 
given fraction is a " proper" fraction and in its lowest terms. If 
the fraction to be integrated is an " improper " fraction, that is, 
if the numerator is of the same or of a higher degree than the 
denominator, the first step is to divide the numerator by the 
denominator. The result of such division will be a polynomial, 
or a polynomial plus a proper fraction. The polynomial can be 
integrated by elementary rules, and the fraction by the methods 
of the present chapter. If the given fraction be not in its lowest 
terms, it must be reduced by dividing out the factors common to 
numerator and denominator. 



/sy*4 /v>3 O 
— 7~ a ; — dx. 
x(x 2 - 1) 



x(x 2 

This is an improper fraction. Dividing numerator by denominator, we 

have 

3 4 - 3 3 - 2 . . x 2 - x - 2 - . x -2 

= x - H -~ — = 3 - 1 + 



x(x 2 - 1) x(x 2 - 1) xix - 1) 

We now separate the last fraction into its partial fractions. To this 
end we assume 

x-2 A . B 

= - + 



x(x — 1) X x — 1 

Clearing of fractions, 

x-2 = A(x- 1) +Bx. 

By substituting x = 0, and x = 1, it is readily found that A = 2, and 
B = — 1, and therefore 

3-2 = 2 1__ 

X{X — 1) 3 3 — l' 

a result which may be readily verified, and the initial assumption thereby 
justified. Hence finally, 



x 



3 3 — 2 , r , ,x , , n rdx r dx 



i^*-Je.-i>* + ajf-J 



= ^ (^ - I) 2 + log 



2 X y °3-l 



402 CALCULUS §§258-259 

258. Exercises. Evaluate the following integrals : * 
4 x — 3 7 r x*dx 



- SiHfh*- 4 / 



x 2 + x - 6 J (a; 2 - 1) (x + 2) 

/• z + 4 , _ r 2 x 2 + 9 a; - 27 , 

2 - I ~i — i — r^ dx - 5 - — , — dx - 

J x 2 — ox -\- o J x 3 — 9x 

o f s<fa . fi f z 3 + * 2 + g V 

d ' J x 2 - x - 12 °- J (x 2 - 1) (x 2 - 4) 

The foregoing method avails to separate a fraction into partial fractions 
when some or all of the factors of the denominator are imaginary. Thus 
- r dx i r dx i r dx i, x + i , . „ / — =\ 

/dx % x ~\~ i 
„ , - = tan -1 x, we have tan -1 x = - log . + constant. 
x 2 + 1 2 x — i 

8 - L* - t\ 2 dx = H^' log( * - l + 0.+4 3 - 1 ' log( * - l -° 

= log Vx 2 - 2 z + 2 + ilog X ~ ] + \ • 

a; — 1 — i 

9. Show also that 

f 2 X t\ o dx = lQ g ^ 2 -2^ + 2 + 2 tan" 1 (re - 1), 

and compare this result with that in 8. 

In cases 3 and 4 will be explained another method of integrating frac- 
tions whose denominators contain imaginary factors. 

259. Case 2. Some of the linear factors of the denominator are 

repeated. 

x z 4- 1 

Example. To evaluate — Tzidx. 

x x — \y 

W§ assume here that 

(l/ * 3 + l = A + _g_ + CD. 

K } X(x~\y * T J! ; -l T (lf- i) 2 (x- 1) 3 

Clearing of fractions, 

(2) x 3 + 1 = A(x - l) 3 + Bx(x - l) 2 + Cx(x - 1) + Dx. 

When x = 0, 1 = -A whence A =— 1. 

When x = 1, 2 = Z> whence D = 2. 

When a; =-1, 0=-8.4-4£ + 2(7-D. . whence2B-C = 3. 

When x = 2, 9 = ^ + 2B + 2C + 2Z). . . . whence £ + C = 3. 

* After separating a fraction into partial fractions, the student should 
always test his work, before integrating, by taking the sum of the partial 
fractions, and noting whether this sum is identical with the given fraction. 



§§260-261 INTEGRATION OF RATIONAL FRACTIONS 403 

Therefore B = 2, and C = 1, and 

x 3 + 1 1 . 2 . 1 . 2 



(3) x(x - l) 3 3 + x - 1 + (x - l) 2 ' (x - l) 3 

Since the sum of the fractions in the second member of (3) is actually equal 
to the first member, our assumption (1) is justified. 
Integrating (3), we have 



/J^^--log^lo g (,-l) 2 -^I-7^ = ^ 



(z-D 



x(x-iy & ■ &v x-\ (x-v 2 ° x (x-iy 

260. Exercises. Integrate the following:* 
- f 2a ; - 5 , A f a: 2 c?a: 

2 r__dx__ ii?m£. Assume the given frac- 

J < 



x{x + l) 2 tion equal to 

3 r dx A B C D 

J X\X + 1)' X -l^( X - 1)2^3; +1 + ^+1)2 

• J (x - 1) 3 dx - 7 - J x 2 (l - xY dx ' 

r 3 a; + 2 f sris 

da; 



9 ■/! 



a; 2 + a 2 ) 2 

261. Integration of a Rational Fraction when the Denomi- 
nator Contains Imaginary Linear Factors. In separating a 
fraction into partial fractions by the methods of cases 1 and 2, 
some of the component fractions will be imaginary whenever the 
denominator of the given fraction has imaginary linear factors, 
and in that case imaginaries will appear in the integral also. We 
have had examples of this. We know, however, by Art. 256, that 
when the denominator has imaginary linear factors, these occur 
in pairs and can be combined into real quadratic factors. We 
shall now show how the given fraction can be resolved into partial 
fractions whose numerators are real, and whose denominators are 
the real linear and quadratic factors of the given denominator. 

* See footnote on page 402. 



404 CALCULUS §261 

Case 3. None of the imaginary linear factors of the denominator 
are repeated. 

7 ^7^ — 7T — r-^dx. 

(x - 1) (a; 2 - 2 x + 2) 

The linear factors of x 2 — 2 x + 2 are imaginary. We assume 

m a; 2 -3 a; + 5 ^ A £a; + (? 

(a;- 1) (a; 2 -2a: + 2) a; - 1 a; 2 - 2x + 2* 
Clearing, 

(2) x 2 - 3 x + 5 a A(z 2 - 2 x + 2) + 5a;(a; - 1) + C(s - 1). 

When x = 1, 3 = A ; when a; = 0,5 = 2i-C, whence C = 1. 

When 3 = 2, 3 = 2 A + 2£ + C, whence B = -2. 
Therefore 

,„x x 2 - 3 a; + 5 = __3 2 a; - 1 

U (a;- 1) (a; 2 -2a; + 2) a; - 1 z 2 -2a: + 2' 

The truth of (3) may be verified by calculation, and this justifies assump- 
tion (1). 

Integrating (3), 



J x — 1 ^ z 2 — 2 i 

2 a; - 2 7 f dx 



/ x 2 — 3 a; + 5 , _ ^ C dx _ r **, 

(x - 1) (x 2 - 2x + 2) aX ~ 6 J x - 1 J x 2 - 2x + 2 aX 



= io g (x-iy-f x2 _ x 2x + 2 dx-f {x _* 2 + 1 



Example 2. To evaluate f — 
J x A 



dx 



+ z 2 + l 

a; 4 + x 2 + 1 = z 4 + 2 a; 2 + 1 - x 2 = (x 2 + l) 2 - x 2 
= (x 2 + x + 1) (z 2 - x + 1). 

The linear factors of both these quadratic factors are imaginary. We 
assume 

(1) 1 Ax + B . Cx + D 
U a^ + ar' + l a: 2 + z + l a; 2 - a; + 1* 
Clearing, 

(2) l = Aa:(a; 2 -a: + l)+5(a: 2 -a; + l) + Ca;(a: 2 + a; + l) + i)(a; 2 + a; + l). 
When a; = 0, B + D = 1; when x = i, 1 = A - iB - C + iD, 
whence A — C = 1, and 5 — 2) = 0. 

Therefore £ = J, D = |. 

When Z = 1, l = A+£ + 3C + 3Z), whence A + 3 C = -I. 

Therefore C =—\ and A = \. 



§262 INTEGRATION OF RATIONAL FRACTIONS 405 

Hence 

1 _1T x + 1 x- 1 -i 

x* + x 2 + 1 2 Lx 2 + x + 1 x 2 - x + 1 J ' 

a result which may be verified by calculation. 

r x + 1 ^ = l r 2x + 1 , ,1 r c?x 

Jx 2 + x + l X 2^x 2 + x+l 2^ 2 + x + l 

= iio g (, 2 +,+i)+- 1 r^+^ 



Similarly, 



2 ■> (x + |) 2 + I 

= ±log (x 2 + x + 1) +^r tan" 1 ?^±1 

2 V3 V3 



x- 1 , 1, , o , *x 1 . ,2i- 1 



^ r 2 — rr 4- 1 2 



tan _: 



x 2 -x + l 2 V3 V3 

Therefore 
C dx 1 , x 2 + x + 1 , 1 H _, 2 x + 1 . , . 2 x - 1] 

1. x 2 + x + l . 1 , . V3 X 

= i ^-s + i + ^71 tan ~ l rrp- 

Observe that after substituting i for x we assume that A, B, C, D are 
all real, an assumption that is justified only by the correctness of the 
result. 

262. Exercises. Integrate the following:* 

1 f x 2 + x + 1 , r 2xdx 

L J ,M4-n dx ' 5 ' J] 



x(x 2 + 1) J (x 2 + 1) (x 2 + 2) 

' x* + x + l , . /■ o: 2 + 2x-l 



2 /SSKr* ° / 



(* 2 + 1) ' J (x 2 + 2) (x 2 + 1) 

Hint. Assume the given frac- « r x 2 — 1 , 
tion to be equal to ^ x 4 + x 2 + 1 

A . B . Cx + D g r 3 x 2 dx 

x x 2 x 2 + 1 ' ' J (x 2 + 1) (x 2 + 4) 

„ r xc?x q r dx 

■ ^ (x + l)(x 2 + l)" ' ^ 1 -x 4 ' 



dx. 



x 4 -3x 2 +l 

* See footnote on page 402. 



406 CALCULUS § 263 

263. Case 4. Some of the imaginary linear factors of the de- 
nominator are repeated. 

/dx 
( 2 i 2W where m is a posi- 
tive integer. 
The formula is 

^V (x 2 +a 2 ) m = 2(m-l)a 2 \_(x 2 +a 2 ) m - l+ ( 2m ~ S ^J (x 2 +a 2 ) m ~\ 

This formula may be verified by differentiating it. It serves to 
reduce the calculation of the given integral to that of the simpler 

/dx 
/ 2 , — 2\m-v anc ^ by repeated application of the formula 
{x -J- a ) 

we arrive ultimately, because m is a positive integer, at the integral 

dx x 

whose value is known to be -tan -1 -, and thus the 



/ 



x 2 + a 2 ' a a' 

given integral is completely evaluated. 

/dx 
( 2_L 1\Z 
(x ~\~ a ) 

In this case m = 3, and by our reduction formula (R) 

r dx 1_ r x . „ /• jg "[ 

J (x 2 + a 2 ) 3 ~ 4 a 2 L(x 2 + a 2 ) 2 "*" J (x 2 + a 2 ) 2 J ' 

Applying the reduction formula to the last integral, we have 

C dx _ J_ r x , r dx 1 = j_ r x , 1 , , xi 

J (x 2 + a 2 ) 2 " 2a 2 U 2 + a 2 Jx 2 + a 2 J 2a 2 Lx 2 + a 2 a a J ' 

Therefore, by substituting and reducing, 

r dx 1 r 3x 3 + 5a 2 x 3 ^Xl 

J(x 2 + a 2 ) 3 8a 4 L(a; 2 + a 2 ) 2 o an a J ' 

/c?x 
(2*» + 4s + 3)«" 

dx _ 1 r d(x + 1) 

(2x 2 + 4x + 3) 2 ~ 4 J [(x + 1) 2 + I] 2 ' 

Applying to the last integral formula (R) with a 2 = \ and m = 2, we have 

r d(x + 1) = g + 1 /■ rffe+jj 

J [(x + l) 2 + i] 2 (x+l) 2 +| + J (x + l) 2 + ± 

= 2x 2 + 4 + x + 3 + ^ 2tan - lV ' 2 ^ + ^ 



§263 INTEGRATION OF RATIONAL FRACTIONS 407 

Therefore 

J (2a; 2 + 4a; + 3) 2 22z 2 + 4z + 3 2 \/2 
II. The general case. 

Example. To evaluate f — . - , ..„ dx. 

J x{x 2 + 4) 2 

We assume that 

m x* + 2x 2 + 5x + 8 _ A Bx + C . Dx + E 

U z(a; 2 + 4) 2 x + z 2 + 4 + (x 2 + 4) 2 ' 

Clearing of fractions, 

(2) x 3 + 2x 2 + 5 x + 8 = A O 2 + 4) 2 + Bx\x 2 + 4) + Cx(x 2 + 4) + Dx 2 + #z. 

When x = 0, 8 = 16 A whence A = £. 

When z = 2 1, -8-i - 8 + 10 i + 8 = 2 1 = -4 D + 2 i#, 
whence D = and Z£ = 1. 

When a; = i, -i - 2 + 5i + 8 = 6 + 4i = 9 A - 3 £ + 3iC -D + %E, 
whence 6 + 4 i = | - 3 B + 3 iC + % , 

f + 3i=-3£ + 3iC, 
and from this, B = — J, C = 1. 

Therefore 

r „x a; 3 + 2a; 2 + 5a; + 8 = 1 r 1 _ a; - 2 , 2 n 

^ j x(x 2 + 4) 2 2 La; x 2 + 4 "^ (a: 2 + 4) 2 J ' 

Our assumption (1), and the tacit assumption that A, B, C, D, E are all 
real, are justified by showing that (3) is a true equation. 
Integrating (3), we have 

/ a; 3 +2a: 2 +5a; + 8 _lr rdx_ r xdx , 2 /• c£r . g /■ cfc ~ | 
a:(x 2 +4) 2 iC ~2!_J a; Jx 2 + 4~ h J z 2 + 4 + J(a: 2 + 4) 2 J' 

By reduction formula (R), 

/ dx _ 1 r x , 1 , _! an 
(a: 2 + 4) 2 "8La; 2 + 4 + 2 tan 2 J ' 
Therefore 

/a: 3 + 2x 2 + 5z + 8 , In x , , . a; , 1 , , a: . 1 x ~| 

,(, 2 + 4) 2 ^ = 2L l0g Va^Tl +tan 2 + 8 tan_1 2 + 4a7+i] 

= ^ T 4 lQ g -7^T7 + 9 tan- 1^ + -^-1 . 
16 L x 2 + 4 2 a; 2 + 4J 



408 CALCULUS §§264-265 

-g— r 7— can always be inte- 

grated, and that no imaginaries will occur in the result so long 
as A, B, p, q are all real. Observe, further, that the methods of 
cases 3 and 4 — that is, the assumption that the numerator of the 
component fraction is of the form Ax + B when the correspond- 
ing factor of the given denominator is a quadratic — is valid even 
when the roots of this quadratic factor are real. 

264. Exercises. Integrate the following : * 

x s -f- x 2 + 2 , . r x z dx 



/ x° ■+■ xr j- z j . r 

(z 2 + 2) 2 ' J{x*+iy 

r x 4 + x 2 -2x - 16 , _ r dx 

J x(x 2 + 4) 2 ' *' 'J x(x 2 + l) 3 

J (3 - l) 2 0z 2 + l) 2 J x(x 2 + l) 3 



4a; 2 + 3x + 2 , 
da;. 



265. The Integral of Any Rational Function. The foregoing 
methods for separating a rational fraction into partial fractions 
were based upon certain assumptions that had to be justified in 
each problem by actual calculation. A general proof can be given 
that these assumptions are always valid, and that consequently 
the methods based upon them always lead to correct results. We 
do not give a proof of this.f 

It follows then that by the foregoing methods any rational proper 
fractipn whatever can be integrated. Since we know also how to 
integrate any rational integral function, that is, any polynomial, 
we can now assert that any rational function whatever, whether 
integral or fractional, can be integrated. And we may assert, 
further, that the integral of such a function is either itself a 
rational function, or the logarithm of a rational function, or the 
anti-tangent of a rational function, or an algebraic sum of such 
functions. 

* See footnote on page 402. 

f For an excellent general treatment of Partial Fractions, which includes a 
proof of the principle stated in the text, see Fine's College Algebra, p. 236 fif. 



266 INTEGRATION OF RATIONAL FRACTIONS 409 

266. Miscellaneous Exercises. Integrate the following: 

/dx r x 2 ~\~ 6 

a7+T 10 ' J 2x* + 7x* + 3x dx ' 



( r x clx /* qj£ 

" J z 3 + l* U * J x(a + bx 2 )*' 

1 /Jtt i2. r * 3+4 



z 2 + l 



+ ar + £ 

L r — *? — , 

J x 2 - 2 a; + 5 



(z 2 + 4z + 5) 2 



<• J^ + z-15^ 13. f-* 

J a: 4 + 



a: 4 + l 
z 2 (z + l)~ Hint. z 4 + l = a; 4 + 2a; 2 + l-2a; 2 



/ x 2 — x — 3 , 
3^ + 1^ 

f" 

J a: 3 + 2a; 2 + a; =(x 2 + V2x + l) 



, 3^ + 31+4 ^ = (x 2 + l) 2 - 2 x 2 



'■f ?*&**■ <*-vs.+d. 

8 -/l 




CHAPTER XXXIII 
ENVELOPES 
267. Systems of Curves. The equation 

in which t is an arbitrary constant to which any value whatever 
can be assigned, represents an infinity of curves, one for each 
value of t. Such an infinity of curves is termed a system or family 
of curves, and t is termed the parameter of the system. 

We have already had examples of families of curves in Chapter 
XX. Other examples are the following. The student should prove 
the statements made in each example. 

1. y = 3 x + t represents all right lines which have the slope 3. 

2. y = tx + 3 represents all right lines which pass through the point 
(0, 3). 

3. (x — t) 2 + y 2 = 4 represents all circles of radius 2 whose centers 
lie on the z-axis. 

4. x cos + y sin d = a, 6 being the parameter, represents all lines 
which lie at the constant distance, a, from the origin. 

5. ■ 1 — ^— = k represents all lines of which the part included 

cos sin 

between the axes has the constant length k. 

/y»2 tip 1 

6. «| — h v = 1 represents all ellipses whose axes are in coincidence 
J t k % 

and'which have a constant area irk. 

The two equations, — + f- = 1 and nab = irk, or ab = k, represent 
a L b l 

together the same family of ellipses. 

7. y = Ix + — j I being the parameter, represents the system of tan- 
gents to the parabola y 2 = 2mx. 

268. Exercises. 

1. Write the equation of the system of tangents to the ellipse, taking 
the slope as the parameter. 

410 



§269 ENVELOPES 411 

2. Write the equation of the family of lines which make with the axes 
of coordinates a triangle of constant area. Draw a figure showing this 
system of lines. 

3. Write the equation of the system of tangents to the parabola, taking 
for parameter the ordinate of the point of contact. 

Oj <yyr qj .3 

4. Show that y -\ y x — -f— - = is the equation of the system of 

m 2 m? 

normals to the parabola y 1 = 2 mx, y\ being the parameter of the system. 

5. Draw a figure of the system of circles (x — t) 2 + y 1 = 4. 

6. Draw several curves of the system y 2 = 4 ax — a 4 . 

7. Draw several curves of the system ty = x l . 



). Envelopes. The tangents to any curve constitute a fam- 
ily of lines having a single parameter; for, the equation of the 
family of tangents can be given in terms either of the slope, or 
of one of the coordinates of the point of contact, and then this 
slope or this coordinate is the parameter of the family. We have 
had illustrations of this principle in the preceding article. Now 
any curve may be regarded as the locus of the intersections of 
consecutive tangents of its system of tangents, and from this point 
of view the curve is termed the envelope of its tangents. We have 
then a system of lines (the tangents), given by an equation con- 
taining a single parameter, and we define the envelope to be the 
locus of the intersections of consecutive lines of the system. 

More generally, in the case of any system of curves given by an 
equation containing a single parameter, f{x, y, t) = 0, the locus 
of the intersections of consecutive curves of the system is defined 
to be the envelope of the system of curves. 

For example, the envelope of the circles, (x — t) 2 + y 2 = 4, is 
plainly the two lines y = ±2. 

Again, consider the system of lines of which the part included 
by the axes is of constant length h. The equation of the sys- 

X li 

tern is + - — - = k. If we draw with some care a number of 

cos 9 sin 

lines of the system, it will be at once apparent (figure on next page) 
that the intersections of consecutive lines constitute a curve which 
looks like the astroid, and which will presently be shown to be the 
astroid. 



412 



CALCULUS 



269 



Our next problem is to de- 
rive from the equation of a 
system of curves 

fix, y, t) -= 
the equation of its envelope. 

Let t and t + h be two values 
of the parameter, and then the 
equations of the corresponding 
curves are 

(a) f(x, y, t) = 0, 

fix, y,t + h) = 0. 
The limiting positions of the intersections of these curves as h = 
are, by definition, points on the envelope. Now the equation 
fix, y,t + h) - f{x, y, t) 




(b) 



= 



represents a curve which passes through all the intersections of the 
two curves of (a). As h = 0, the limit of (b) is, 
dfjx, y, t) 
dt 



0. Hence fix, y, t) 



an d M^M =0 



dt 



intersect in those points of the envelope which lie upon/(x, y,t) =0. 
Therefore to obtain the equation of the envelope we have only to 
eliminate t from the equations 



(c) 



fix, y, t) = and 



dfjx,y, t) 
dt 



0. 



Examples. 

1. Let us find the envelope of the system of parabolas y 2 = 3tx — i 3 . 
In this case 

fix, y, t) = y 2 - 3 tx + V = and f£ = -3 x + 3 f = 0, 

dt 

whence t 2 = x, t=± Vx, and on substituting in the equation of the 

system we have 

y 2 = ±Vx (3x — x) = ±2x* and y A = 4:X z , 

which is the equation of the envelope. 

2. Let us find the equation of the envelope of the system of lines from 
which the axes of coordinates cut off the constant length k. As we saw 
in Art. 267, example 5, the equation of this system of lines is 

JK ' y ' cos sin 



-k, 



§ 270 ENVELOPES 413 

Taking the partial derivative as to 0, 

d/(z, y, e) s xsrnd _ ycosO = Q whence _V__ _. zsin 2 
3d cos 2 6 sin 2 ' sin0 cos 3 

Substituting in the given equation, we get 
x cos 2 d -\- x s in 2 
cos 3 

whence z = & cos 3 0, ?/ = A; sin 3 0, 

and these are the parametric equations of the envelope, which is seen to be 
an astroid. 

We shall now solve this problem by another method which is pref- 
erable in many cases. 

The system of lines may be given by the two equations 

- + 1 = 1 and a 2 + b 2 = k 2 , 
a o 

and a will be regarded as the parameter. Then 

a b da a 2 b 2 da 

To find — we differentiate a 2 -f- b 2 = k 2 , as to a, and get 
da 

i z. db A , d& a 

a + 6 — - = 0, whence — = — - • 
da da o 

Substituting in ~ = 0, and reducing, we have 
da 

i£ — K y b 2 x 

a 3 b 3 b a 3 

x u 
Substituting in the given equation, - + 1 = 1, we have 

a o 

x . b 2 x ^ a 2 -\-b 2 1 k 2 x 1 , x y 1 

- + — = 1, — x = 1, — = 1, and - ■ = g - = - - . 

a a 3 a 3 a 3 a 3 6 3 A: 2 

From these last equations, a 2 = &w, 52 _ ^3^ and on substituting in 
a 2 -J- 6 2 = ft 2 , there results finally 

xl+y% = k*, 
and this is the equation of the envelope. 

270. An Important Property of the Envelope. This property 
is, 

The envelope is tangent to each curve of the system. 

Proof. Every point of the envelope is a point on some curve 
of the system, and what we have to show is that at such a point 
the two curves have the same slope. 



414 CALCULUS § 270 

The slope of the curve j{x, y, t) = at the point (x } y) is given 
by the equation 

dx dy dx 
The equation of the envelope is obtained by eliminating t from 

the equations / = 0, -^ = 0. Let us suppose that this elimina- 
ot 

^ /• 

tion is performed by solving— = f or t and substituting this value 

Ov 

for t in / = 0. Then the equation of the envelope is 

(b) f(x,y,t)=0, 

where t is no longer a constant but a function of x and y. To find 
the slope of the envelope at the point (x, y) we may therefore 
differentiate (b) as to x, regarding t as a function of both x and y. 
This gives 

( C ) # + #& + # *_o 

y ' dx ^ dydx^ dt dx 

df 

If we suppose that the value of t obtained by solving -— = to 

ot 

be t = (f>(x,y), then 

,,. dt _ d(f> &4> dy 

dx dx dy dt 

And by substituting this value for -=- in (c) we get the equation 

dv 
which gives the slope, -p, for the envelope. But it is unnecessary 

to make this substitution because — = 0, whereupon the last term 

ot 

of (c) drops out and (c) becomes 

df v^ = 0< 

dx dy dx 

dy 
This, which gives the slope, -p, for the envelope, is identical with 

(a), which gives the slope for the curve / = 0. Hence the curve 
and the envelope have the same slope at the point (x, y), and this 
was to be proved. 



§ 271 ENVELOPES 415 

271. Exercises. Find the equations of the envelopes of the 
following systems of curves. For a thorough understanding of the 
problem it is essential that the student draw a figure, in each case, to 
show roughly the family of curves and the envelope. 

1. The system of parabolas, y 2 = 4 t(x — t). 

2. The system of parabolas, y 2 = 4 tx — t 4 . 

3. The system of lines, y = Ix + — • 

4. The system of lines which lie at the constant distance, a, from a 
fixed point. 

5. The system of cubics, y 2 = 4 (x — t) 3 + 3 t. 

6. The system of exponential curves, y + t = e*^ 1 . 

7. The system of logarithmic curves, ty = log(x + t). 

8. The system of sine curves, y + t = sin(z + t). 

9. The system of curves, ty = xK 

10. The system of lines that make with the axes of coordinates a 
triangle of constant area. 

11. The system of circles whose centers are on the parabola, y 2 = 2 mx, 
and which are tangent to the 2/-axis. 

12. The system of circles whose diameters are the double ordinates of 
the parabola, y 2 = 2 mx. 

13. The system of ellipses whose axes are in coincidence and which have 
the constant area irk (see Art. 267, example 6). 

14. On chords through the vertex of the parabola y 2 = 2 mx, as diam- 
eters, circles are described. Find the polar equation of the envelope. 

15. From the origin lines are drawn terminating in the hyperbola 
xy = 2 a 2 . On these lines, as diameters, circles are described. Find the 
polar equation of the envelope of these circles, and show that it is a 
lemniscate having the axes of coordinates for flex- tangents. 

16. Circles are drawn through the origin with their centers on the hyper- 
bola xy = 2 a 2 . Show that the envelope is a lemniscate similar to that 
of exercise 15. 

17. Circles are drawn through the origin with their centers on the as- 
troid £3 + y* = cfi. Find the polar equation of their envelope. 

18. Through a fixed point on the circumference of a circle chords are 
drawn, and on these as diameters circles are described. Show that their 
envelope is a cardioid. 

19. Find the envelope of the normals of a parabola. (See Art. 268, 
exercise 4.) Compare example 1, Art. 110. 



416 



CALCULUS 



§272 



20. From the points where the normals of a parabola cut the x-axis, 
lines are drawn perpendicular to these normals. Show that their envelope 
is a parabola. 

21. In Art. 153, equation 6, we have the equation of the path of a 

projectile „_ w .* - ^ ;**»*.*•. 

Regard 4> as the parameter and find the envelope of all the paths of con- 
stant initial velocity v . Find then the polar equation of this envelope, 
and compare the result with equation 10 of Art. 153. 

22. In exercise 19 above it was found that the envelope of the normals 
of the parabola is the evolute of the curve. That this is true of all curves 
we already know, but it may be proved again in the following way: 
The equation of the system of normals to any curve is 

/ ss x - xi + (y - y x )yx = 0, 
where x x may be regarded as the parameter. Differentiate this equation 
as to Xu remembering that both y x and y x are functions of Xi. Then 
solve the two equations, / = and -— = for x and y, and thus obtain 

OX x 

the parametric equations of the envelope of the normals, which will be 

found to be identical with equations IV of Art. 109. 

272. Caustics. When rays of light, emanating from a fixed 

point, fall upon a concave mir- 
ror, the reflected rays envelope 
a surface that is termed a caustic. 
We shall suppose the mirror to 
be a hemisphere or the half of a 
circular cylinder. Moreover, we 
need consider only a cross sec- 
tion made by a plane through 
the source of light and through 
the center of the sphere or per- 
pendicular to the axis of the 
cylinder. Then all the rays con- 
sidered lie in this plane and the 
caustic is a plane curve. 

We shall take the case where 
the illuminating point is at an 
infinite distance, so that the 




§272 



ENVELOPES 



417 



incident rays are parallel. Let i be an incident and r a reflected 
ray. Then by a principle of optics i and r make equal angles, 9, 
with OP, the normal to the surface at P. 
OP is in this case, of course, the radius of 
the circle. The normal equation of r is 

x cos a + y sin a = p. 
But 

a = — I x — 2 0), cos a = sin 2 0, 
sin a = —cos 2 0, p = asmd, 

where a denotes the length of the radius. 
Hence the equation of r is 

(a) x sin 2 — y cos 2 6.= a sin 0. 

This is the equation of the system of reflected rays, being the 
parameter, and it is the envelope of this system of lines that we 
are to find. Differentiating as to 6, we have 




(b) 



x cos 2 d -\- y sin 20 = - cos 0. 



Multiplying (a) and (b) respectively by sin 2 and cos 2 and 
adding, and then multiplying by cos 2 and sin 2 and subtract- 
ing, we have 



(c) 



x = ^ (2 sin sin 2 9 + cos cos 2 0), 

Li 



- (cos sin 2 — 2 sin cos 2 0), 



And these are the parametric equations of the caustic. Equa- 
tions (c) can readily be reduced to 

(d) x = ? (1 + 2 sin 2 0) cos 0, y = a sin 3 0, 

and these can be brought to the form 



(e) 



x = -r(3cos 6 — COS 3 0), 
y = | (3 sin - sin 3 0). 



418 CALCULUS • §273 

If now in equations (a) of Art. 94 we set a = - , b = j, equations 

(e) will result, and this shows that the caustic is an epicycloid of 
two cusps. 

When the reflecting mirror is a cylinder, the caustic surface is 
also a cylinder, obtained by erecting perpendiculars to the plane of 
the caustic curve at each point of the curve; and when the mirror 
is a hemisphere, the caustic surface is obtained by revolving the 
caustic curve about the initial line. 

273. Exercises. 

1. Show that the total length of the caustic curve is 3 a. 

2. Show that the area inclosed between the caustic curve and the circle 
is J ira 2 . 

3. Find the area of the caustic surface when the mirror is a hemisphere. 

4. The reflected rays which envelope the caustic fall a second time 
upon the mirror and are again reflected. This second system of reflected 
rays envelopes a curve called the second caustic. Show that its para- 
metric equations are 

— x = - (4 sin sin 4 + cos cos 4 0), 

y = - (4 sin cos 4 — cos sin 4 0). 

5. When the illuminating~point lies on the circumference of the circle, 
the reflected rays envelope a curve also termed a caustic. Show that its 
parametric equations are 

x = ^ (3 sin sin 3 + cos cos 3 0). 

— y = - (3 sin cos 3 — cos sin 30). 
o 

6. When the reflecting surface is the cycloid 

x = a (0 — sin 0), y = a (1 — cos 0), 
and the incident rays are parallel to Y, show that the resulting caustic 
is also a cycloid. 

7. When the reflecting surface is the logarithmic spiral p=ae be , and the 
source of light is at the pole, show that the caustic is also a logarithmic 
spiral. 

8. Let the reflecting surface be any curve whatever, and let the incident 
rays be parallel to OX. If (x,y) be the point of incidence on the given 



§ 273 ENVELOPES 419 

curve, and if (a,/3) be the corresponding point on the caustic, show that 
the parametric equations of the latter curve are 

where x is the parameter and y, y', y" may be expressed in terms of x by 
means of the equation of the given curve. 

Use the equations of exercise 8 to find the equations of the caustics of 
the following curves. 

9. The parabola y 2 = 2 mx. 12. The equilateral hyperbola 

10. The circle of Art. 272. 9 xy = 4. 

11. The ellipse 13. The hyperbola 

x — a cos 0, y = b sin 0. x = a sec 6, y = b tan 0. 

14. The cycloid generated by a circle rolling on the ?/-axis. Compare 
exercise 6. 



BOOK VI 

AN INTRODUCTION TO 
ORDINARY DIFFERENTIAL EQUATIONS 



CHAPTER XXXIV 

ORDINARY DIFFERENTIAL EQUATIONS OF THE FIRST 

ORDER * 

274. Definitions. A differential equation is an equation that 
contains derivatives or differentials. The following are examples 
of differential equations, many of which have occurred in the 
preceding chapters: 

2 d?y &y<± 16zcos2z. 
dx z dx* 

3. a&=y Art. 145. 

dx 

4. p dd = tan </> dp Art. 97. 

5. — - = — a sin /d Art. 147. 

CLt 

6 - 4 1+ (l) 2 ]=^ or l=4 v ^ 2 Art - 145 - 

* In this and the following chapter are explained the methods of solving 
those classes of ordinary differential equations that the elementary student 
of applied mathematics is most likely to meet with. Those who wish a more 
extensive knowledge of differential equations are referred to the textbooks 
on that subject. Those to be especially commended are An Elementary 
Treatise on Differential Equations, by Dr. A. Cohen of The Johns Hopkins Uni- 
versity (D. C. Heath and Co.), and Introductory Course in Differential Equa- 
tions, by Dr. D. A. Murray of McGill University (Longmans, Green and Co.). 

421 



422 DIFFERENTIAL EQUATIONS §275 

7. L } dx)A = r Art. 109. 

dx 2 

8. d 2 = ^F Art.151. 

dt 2 s 2 

9. {y - z)^-+ (z - x)~ = x - y Art. 208. 

dx By 

10. fl + Tl + w = Art. 202. 

dx 2 dy 2 

In the preceding chapters many other differential equations 
have been met with. 

A differential equation that contains only one independent 
variable, and therefore only ordinary derivatives, as distinguished 
from partial derivatives, is termed an ordinary differential equa- 
tion. 

A differential equation that contains more than one independent 
variable contains also partial derivatives, and is called a partial 
differential equation. 

Of the above examples, the first eight are ordinary and the last 
two are partial differential equations. 

We shall consider in this book only ordinary differential equations. 

The order of a differential equation is the order of the highest 
derivative in it. Of the examples above, 3, 4, 5, 6, and 9 are of 
the first order, 1, 7, 8, and 10 are of the second order, and 2 is of 
the third order. 

The degree of a differential equation is the exponent of the 
highest power of the derivative of highest order in it, after the 
equation has been so transformed by clearing of fractions and 
rationalizing that neither the dependent variable nor any of its 
derivatives occur in the denominator of a fraction or under a 
radical sign. 

Of the examples just given, all are of the first degree, except 
6 and 7, which are of the second degree. 

275. Solution. Particular Integrals. The Complete Primi- 
tive. A solution of a differential equation is any relation among 



§275 EQUATIONS OF THE FIRST ORDER 423 

the variables that satisfies it. Consider, for example, the differen- 
tial equation 

■ , d 2 y dy . 

The relations y = x and y = e x both satisfy this equation; for, 

from y = x, we have -p = 1 , -p^ = 0, and on substituting in the 

given equation there results 

(x—.l)Q'—x + x = 0. 

dy d 2 y 

And, from y = e x , we have -p = e*, -p^ = e*, and on making these 

substitutions 

(x - l)^ - xe x + e* = 0. 

Moreover, the given equation is also satisfied by the relation 
y = C\X + c 2 e x , where C\ and c 2 are arbitrary constants; for on 

dy d 2 y 

differentiating this we have -p = Ci + c 2 e x , -p| = c 2 e a: , and putting 

these values in the given equation we get 

(x — 1) c 2 e x — e(ci + c 2 e x ) + Cix + c 2 e x = 0. 

Now it is a fact that ?/ = C\X + 026* is the most general solution that 
the equation can have. It is called the complete solution or the 
complete primitive of the given differential equation. The solu- 
tions y = x and y = e x are special cases of the complete primitive, 
and may be gotten by giving particular values to the constants 
C\ and c 2 . They are termed particular solutions or particular 
integrals. All solutions derived from the complete solution by 
giving particular values to c\ and c 2 , such as 2 x — 3 e x , are par- 
ticular integrals. They are plainly infinite in number. 

The foregoing definitions apply to every differential equation. 
The most general solution is termed the complete primitive, and 
every special case of this is a particular integral. 

Every problem of integration, like those of the preceding chap- 
ters, is in fact a problem in solving a (very simple) differential 

equation. For instance, to integrate J cosxdx is the same thing 



424 DIFFERENTIAL EQUATIONS §276 

dy 
as to solve the differential equation ~f-= cosz, or dy = cosxdx. 

This fact, and the further consideration that to solve a differential 
equation is to pass from an equation that contains derivatives to 
one that does not contain them, would lead us to suspect that the 
process of solving a differential equation may involve one or more 
processes of integration. Now every integration introduces an 
arbitrary constant, and we may therefore expect the complete 
primitive to contain arbitrary constants. And such is in fact the 
case. Further, the complete primitive of a differential equation 
of the first order has one arbitrary constant, of one of the second 
order two arbitrary constants, and so on. The general principle, 
the proof of which is given in the following article, is as follows: 

The most general solution, that is, the complete primitive, of an 
ordinary differential equation of the nth order, contains n arbitrary 
constants, and no more. 

276. Derivation of the Differential Equation from the Com- 
plete Primitive. Let us find the differential equation whose com- 
plete primitive is 

y = CiX z + c 2 x. 
Differentiating this, 

dx~ SClX +C2 ' dx*~ bClX ' 

Solving these last two equations for C\ and c 2 , we have 

1 d 2 y dy x d 2 y 

6x dx 2 2 dx 2 dx 2 

Substituting these values of c\ and c 2 in the given equation and 
reducing, there results 

Starting here with an equation in x and y which contains two 
arbitrary constants, we differentiate twice, eliminate the constants 
from the three equations, and arrive at a differential equation of 
the second order. 

More generally, if we start with an equation in x and y that con- 



§§277-278 EQUATIONS OF THE FIRST ORDER 425 

tains n arbitrary constants, an equation free from those constants 
may be obtained in the same way as in the above example. That is, 
we differentiate n times and have then n -\- 1 equations from which 
the n constants may be eliminated. This may usually be done by 
solving some n of the equations for the n constants, and putting 
the values so obtained in the remaining equation. Of course any 
method of eliminating the constants is allowable. The result 
will contain derivatives of the nth and usually of lower orders, 
and is therefore a differential equation of the nth order. 

The truth of the principle of the preceding article, that the 
complete primitive of a differential equation of the nth order has 
n arbitrary constants, is at once apparent; for if the complete 
primitive had fewer or more than n constants, then the differ- 
ential equation derived from it by the process explained above 
would be of an order different from n. 

277. Exercises. Find the differential equations of which the 
following are the complete primitives : 

1. y=c 1 x*+c i . ' 7. {x - Cl y + (y - c 2 ) 2 = r 2 . 

2. y = cx + Vl- c\ 8. x 2 + y 2 + 2c 1 x + 2c 2 y + c 3 = 0. 

3. y = c lX + c^ + x\ Q y = Cl x + C ->- 

4. y = Ci sin x + c 2 cos x. x 

5. y = c x e x + c 2 e~ x . 10 y = CiX + £2 + c ^ 

6. y = Cl e x +c 2 e 2X +c z e 3X . x 

278. Geometrical and Physical Interpretation. The complete 
primitive represents a system of curves (Art. 267) of which the 
differential equation expresses the characteristic geometric prop- 
erty. For example, y = ex 2 represents a system of parabolas. 
Through every point of the plane passes one curve of the system, 
and but one (with the exception of the origin). The differential 

dy 
equation derived from y = ex 2 is#-y- — 2 y = 0, and this expresses 

the property that the slope at the point (x, y) of that particular 
curve which passes through that point is equal to twice the ratio 
of the ordinate to the abscissa of the point. The problem of 
finding the complete primitive of this differential equation is 



426 DIFFERENTIAL EQUATIONS §279 

identical with the geometrical problem of determining the system 
of curves of which the slope at any point is twice the ratio of the 
ordinate to the abscissa. We had many such problems in Chapter 
XX. See in particular Art. 146, exercise 2. 

Again, y = C\X 2 + c 2 represents a system of parabolas. Through 
every point of the plane passes an infinity of curves of this system. 

d 2 v dv 
The differential equation derived from this equationis x-~ — -j- = 0, 

and expresses the property that the ratio of the first and second 
derivatives at the point (x, y) of every curve that passes through 
that point is equal to the abscissa of the point. The problem of 
finding the complete primitive from the differential equation is 
identical with the geometrical problem of determining the system 
of curves such that the ratio of the first and second derivatives 
at any point is equal to the abscissa of the point. 

Or this complete primitive and its differential equation may be 
written in another familiar notation, 

d 2 s ds 

where s is the distance traversed by a moving body in time t. 
The problem of deriving the complete primitive from the differ- 
ential equation is identical with the mechanical problem of deter- 
mining the distance traversed by a body when the ratio of the 
velocity to the acceleration is equal to the time. For other prob- 
lems of this kind see Chapter XXI. 

Anil in general, every differential equation can be interpreted 
as expressing a geometrical property common to all the curves 
of a family, or a mechanical property of the motion of a body. 

These interesting geometrical and physical properties of differ- 
ential equations are fully treated in the textbooks on differential 
equations. 

279. Solving a Differential Equation. In the Integral Cal- 
culus we learned that, although every function can be differen- 
tiated, there is no general rule for performing the inverse process 
of integration (see Art. 122). 



§§280-281 EQUATIONS OF THE FIRST ORDER 427 

And so here, although the differential equation can always be 
gotten from a given primitive, there is no general method of per- 
forming the inverse process. That is, there is no general method, 
applicable to every case, of finding the complete primitive when 
the differential equation is given. There are, however, a large 
number of devices that are effective for solving certain classes of 
equations. In this and the following chapter some of the more 
important of these devices are explained. 

280. The Differential Equation of the First Order and Degree. 

This equation is of the form 

2 -**»>• 

or, as it is usually written, 

M dx + N dy = 0. 

M may be a constant, or a function of both x and y, or of one of 
them only; and the same is true of N. By the principle at the 
end of Art. 275, the complete primitive will contain one arbitrary 
constant. 
We shall consider a few integrable cases of this equation. 

,281. Variables Separable. When the equation can be brought 

to the form 

<t>(x) dx + \l/(y) dy = 0, 

where <j>(x) does not contain y, and \p(y) does not contain x, the 
variables are said to be separated. When in this form the equa- 
tion can be solved, for, by direct integration, we get 



J<f>(x) dx + JVfo) dy = 



The problem of solving this differential equation is thus reduced 
to a problem of the Integral Calculus, viz., that of integrating 
functions of a single variable. Such integration is termed quad- 
rature (Art. 160). The equation is regarded as solved even if we 
are unable to carry out the integration indicated. And in the 
case of any differential equation whatever, when the process of 



428 



DIFFERENTIAL EQUATIONS 



§§ 282-283 



solving has been brought to one of quadratures, the differential 
equation is regarded as completely solved, whether it be possible 
to perform the integrations or not. 

Example. x 2 y dx + (1 — x) (1 — y 2 ) dy = 0. 
Dividing by y(l — x), 



dx 



1 



.£*,_ 



dy = 0, 



1 - x y 

and the variables are separated. Integrating, 

/x 2 /*1 — y 2 
- — — dx + *- dy = c, 
1 — x J y 

or 

f [-*-!+ Yh) dx+ f{l- y ) dy = c ' 

and on carrying out the integrations and reducing we have 



log 



y 



= c + x+-{x 2 + y 2 ), 



1 -x 
and this is the complete primitive. 

282. Exercises. 

1. x dy ± y dx = 0. 

2. x 2 % + y 2 dx = 0. 

3. xdy + 2/ 2 cte = 0. 

4. cot dd<j> + cot <f>dd = 0. 

5. x vl + x 3 cfo/ + 2/ das = 0. 

6. sin y dy + sin x cos y dx = 0. 

J 
288. Variables Separable by Transformation. Sometimes 

when the variables are not separable they become so after a 

simple transformation. 

Example, xdy — y{\ — x 2 y) dx = 0. 

The variables are not separable. Set y = vx; then dy - v dx + a;<to, 
and on substituting these values our equation becomes 

x(v dx + x dv) — vx(l — vx z ) dx = 0, 



7. 


dy = 

dx 


e y ~ x . 






8. 


du 
dv 


1 + it 2 






1+v 2 




9. 


2/(1- 


- x) dx — 


x(l - 


y)dy 


10. 


y 


J- 9 ^ In 


dy 
gV dx 


= 0. 


i + 


X 



which reduces to 



x 2 dx = 0, 









\--x 3 = c. 

v 3 






- and reducing, we have 

X 






x z = c, or x(x 2 y - 

y 


-3) = 


= cy, 



§§284-285 EQUATIONS OF THE FIRST ORDER 429 

and the variables are separated. Integrating, 

1 , 1 



Substituting v 



which is the complete solution. 

In Art. 287 will be found a class of equations in which the vari- 
ables are always separable by a transformation. 

284. Exercises. 

o X J V 7 ? { \ 7 ^ ^ = °n S. (x + y tan-i y -dy) 

2. {2xy 2 + y)dx-xdy = 0. V x b I 

3. x -r = x — (y — x) 2 . — [y — x tan -1 - ) dx = 0. 

dx V X J 

Set y — x = ». Set z = P cos d, y = P sin 0. 

4. (a; + 2/) 2 ^ = a 2 - 9 . x dy + [fe _ ^f _ ^ = . 

5. (z 3 vV + ^-^Vl + x 2 ) dx Set 7/ - x = y 2 . 

+ x?/Vl + z 2 d?/=0. 10 # = 

fi dw _ 1 + ^ 2 w "^ 

' dy 1 + y 2 v 2/ a/1 — £ 2 — y 2 -\-x-\fx 2 + ?/ 2 

Set m = tan a, y = tan /?. zVl —x 2 — y 2 — y \/x 2 + y 2 

7. ?/ 2 Vl - 2/ 2 do: Set x = P cos 0, y = P sin 0. 

+ x 2 Vl - z 2 dy = 0. 
Set x = sin a, y — sin p. 

285. The Integrating Factor. Consider the following total 
differentials : 

, , , x dy — y dx 7 y y dx — x dy , x 

xdy + ydx = dxy; — ^—^ — = d-: - = d-\ 

u u u ' x 2 x' y 2 y' 

xdy — ydx dy dx 7 . y 7 7 1 7/ „ , ox 

— - — = — - — = dlog-; xdx±ydy= ^ d(x 2 ± y 2 ) ; 

xy y x x 2 

xdy -ydx ^y 

xdy -ydx _ x 2 _ x _, ^, 



430 DIFFERENTIAL EQUATIONS 

A A dU 

x Vx 2 — v 2 L /v\ 2 x 



2/2 \A-© ! 



A total differential is also called an exact differential. Now, 
when the variables of a differential equation are not immediately 
separable, it is sometimes possible, after multiplying by some sim- 
ple factor such as— 5, -5, — , —z—. — 5, , -, and the like, 

x 2 ' y 2 ' xy' x 2 + y 2 ' Vx 2 - y 2 

to rearrange the terms so that the equation shall be a sum of total 
or exact differentials identical with or similar to those just given. 
The equation itself is then said to be exact, and the factor that 
makes it so is termed an integrating factor. When an equation is 
exact it can always be solved.* 



Example 1. xdy- 


- y dx + x log x dx = 0. 




Multiplying by — , 


we have 






xdy — y dx . , dx 
" * +logz = 

X 2 X 


= 0, 



or 



<(9+i*.?-« 



and the equation is seen to be exact. Integrating, 

y -+±(io g xy = c, 

and this is the solution sought. The integrating factor is - 

x- 



]xi: 



Example 2. (x 4 — x 2 y 2 — y) dx + {xy z — x 3 y + x)dy = 0. 
Writing this in the form 

x(x 2 — y 2 ) (xdx — ydy) + xdy — y dx = 0, 



and dividing by x vV — y 2 , 

* It is proved in books on differential equations that every equation of 
the first order and degree has an integrating factor — has in fact an infinity 
of them. And methods of finding the integrating factor are known for a 
few classes of equations, but there is no general method of finding it that is 
effective with every differential equation of the first order and degree. The 
method of the text is termed "the method of determining the integrating 
factor by inspection." 



§§286-287 EQUATIONS OF THE FIRST ORDER 431 

we have \/x 2 — y 2 (xdx — y dy) -\ y = 0, 

x Vx 2 - y 2 

or 1 Vx 2 - y 2 d(x 2 - y 2 ) + d sin- 1 ^ = 0. 

The equation is the sum of two exact differentials and is therefore exact. 
Integrating, we get as the complete primitive 

\ (x 2 - 2/ 2 ) 1 + sin- 1 ^ = c. 
6 x 

In this case the integrating factor is • 

x Vx 2 - y 2 

286. Exercises. 

i dy ■ 1 9. (x z + xy 2 + x)dy 

dx^ y vi^¥ 2 ' + w + y* 2 -y)dx = o. 

2. xdy — ?/(l — xy) dx = 0. 10. <f> cos <j> d<t> + cos d dd 

3. x z ^ + x 2 + y 2 -x 2 y = 0. = - sin0d0 + ^sin 0<fa. 

dx 

4. (2 x?/ 2 + «/)dx — x dy = 0. 11. x(y 2 cos y — 1) dy 

5. 2/(2 x?/ + e z ) da; - e x dy = 0. + 2/(2/ sin y + l)dx = 0. 

6. w(l - v Vv^^u 2 ) dv 12 <k = J^lJL 



-v(l+vVv 2 -u 2 )du = 0. 



dx x(y — 1) 



7. 2ax 2 2^+(^V-2a^ 2 )dx = 0. 13 - (y 2 - ^ V 1 - 2/ 2 ) 



+ x 2 y -\-x)dy + 2/ V 1 -y 2 dx = 0. 

+ (x 3 + xy 2 — y) dx = 0. 14. (sin y — x cot y) dy + dx = 0. 

287. The Homogeneous Equation. The differential equation 
Mdx + Ndy = 
is said to be homogeneous if M and iV are homogeneous functions 
of the same degree in x and 2/- 

In the homogeneous equation the variables are not separable. 
But by the transformation y — vx, the homogeneous equation 
can always be changed into one in which the variables are sepa- 
rable. The proof is as follows : 

In Art. 249, f(x, y) was denned tc^be homogeneous if 

f(tx,ty) =t"f(x,y). 

In this let t = -, and there results 
x 



f 



(l.9- >">■ 



432 DIFFERENTIAL EQUATIONS §287 

A*,v)-**(i, g-**©- 

Expressed in words, a homogeneous function of the nth degree in x 
and y is equal to x n times a function of - • 

We now write our differential equation in the form 

dy _ _M 
dx ~ N' 



and, because M and N are homogeneous and of the same degree, 

^7 is a functic 

N 

equation becomes 



-j-j is a function of - alone. Calling this function <M-), our 

i\ ■ X \X/ 



dy = (y\. 

dx \x) 
Now set y = vx; ~ = v + x -p » - = v, and our equation becomes 

which can be brought to the form 

dx dv 



x <f>(v) — v 

and the variables are separated, and this is what we were to prove. 
Solving the last equation, 



dv 

J <£M - v 



x = I — 



After the integration has been performed, - is to be substituted 

X r£l 



for v and the result will be the complete primitive of the given 
equation. 

We see, then, that the homoJ^eous differential equation of the 
first order and degree is oneKi which the variables are always 
separable, and therefore one that can always be solved. 



Example, xy dx — (x 2 — y 2 ) dy = 0. 

Set y = vx; then dy — v dx + x dv, and the equation becomes 
x 2 v dx — x 2 {l — v 2 ) (v dx + x dv) = 0, 



§§288-289 EQUATIONS OF THE FIRST ORDER 433 

which reduces to 



dx . 
x 



\V V 6 ] 



and the variables are separated. Integrating, 

log x + log v + — = \ log c, or Ox) 2 = ce v \ 

— v 

Putting - in place of v, we have 
x 

y 2 = ce v 2 
as the complete solution. 

288. Exercises. The student should try the method of 
Art. 285 as well as the method for homogeneous equations. 

1. y 2 dx + (x 2 - xy) dy = 0. 9. xy dx - {x 2 + xy + y 2 ) dy = 0. 

2. ydx-\-{y — x) dy = 0. 

3 dy ^ y 10. log^dx + -dy = 0. 
dx y — x y 

4. xdx+ (y -x)dy = 0. n dy x = 2 

5. zf + 3 y = 4z. ■ dx y ■ 

dz 19 ds t*s + ts 3 

6. {u 2 + v 2 ) dv -uvdu = 0. lJ " dt p + £2 S 2 + s 4* 



7. (^ + Vx 2 -2/ 2 )dx-xd?/ = 0. 



13. (x 4 + 3x 3 ?/)<fy 



8. fz + 2/tan^W + (^ + 3 xy 3 ) dx = 0. 

. y , A 14. (x 4 — x%) dx 

— xtan^cfy = 0. v , /, , x , 

s + G/ 4 - V z x) dy = 0. 

289. The Linear Equation. The linear differential equation 
of the first order and degree has the form 

(a) £ + *-«, 

where P and Q may be constants or functions of x, but do not 
contain y. 

To solve this equation we first mj^ply it by er Pdx , 
Observe now that 



d_ 
dx 



y j™ = /«.g + ,/^p _ /^g + ^ 



434 DIFFERENTIAL EQUATIONS §290 

Therefore 

This equation is exact, and we see that e^ Pdx is an integrating 
factor of (a). On integrating we have 

(b) yeS Pdx = fQeS Pdx dx + c, 

and this is the complete primitive of (a). Therefore the linear 
equation can always be solved. 

Example. -~ — -y = 3x. 

dx 1 — x 2 

This is linear and may be solved by formula (b) . In this case P = — » 

1 — x 1 

fpdx = ff^§ = log vr^, 

and ' / Pd W ogV ^=VT^. 

Also, fQef Pdx dx= f3xVT^tfdx = -(l-x 2 )i+c. 

Hence, on substituting these values in (b) , we have 
y Vl-z 2 = -(1 - x 2 )i+ c, 
which is the solution sought. 

290. Exercises. Note that in some cases' some other method 
of solution is simpler than that of the preceding article. 

tJdy. 1 a ds . 2 st 1 /, — 

V. x-f + y= - ■ 6. -r + = V 1 - t 2 . 

dx vi — x 2 dt 1 — t 2 

2. x^ + 3t/ = 4x. 7. cotx^ + y = m. 

3. {l-x 2 )^-xy = 3x(l-x 2 ). 8 - cosxdyH- 

dx * (ysmx — m cos 2 x)dx = 0. 

4. f* + -^- = 1-x 2 . 9. (l+a/?)da-(l-a 2 )d0 = O. 
dx 1 — x 2 

5. w + zx^-y-o. ' la £ + «»•<** -«** 

ax 

Hint. Consider x the dependent H« d0+ tan <*> d<£ = tan <?> sin tf> d</». 

and y the independent variable. 12. du-{- {u — 1) cos v dy = 0. 



§291 EQUATIONS OF THE FIRST ORDER 435 

13. VT^tf f* + y = 1. 16. ^ - my = nx n ~H m *. 

dx ax 

14. ^ + (y _ C o S X ) sin x - 0. 17 - ^ + 2 *V = e-^cos x. 

15. ylpgydB + (a;-2^)dy = 0. 18 ' *^ ~ y{X + x) = * m 

291. Equations Reducible to Linear Form. Sometimes a non- 
linear equation can be changed into a linear one by a transforma- 
tion of one or both variables. Such is the case with the following, 
known as Bernoulli's equation: 

(a) f x + Py = Qy", 

where P and Q do not contain y, and n is any number. 
Dividing this equation by y n , we get 

y- n i + p y~ n+1 =Q- 

Setting y~ n+l = v, we have y~ n ~f- = __ j~i and our equation 
reduces to 

jg + (l-n)P*-(l-n)Q, 

and this is linear and may be solved by the method of Art. 289. 

Example, x -&■ — y = y 2 log x. 
ax 

We first divide by x and get 

dx x x 

which is of type (a). Putting y~ 2+1 = v, or y = -, we have -& = — « "T". 

v dx v 2 dx 

and therefore 

ldv_ _ _1 _ 1 log a? 
v 2 dx vx v 2 x 
and 

dv_ . v_ _ _ log a; 
dx x x 

which is linear. Solving it, we have 

/» 
v = 1 — log x + - • 
a; 



436 DIFFERENTIAL EQUATIONS §§292-293 

Therefore the complete primitive of the given equation is 
- = 1 — log x H 



292. Exercises 

dx ~ a ' * dx ' 1 + x 



1. (l -0^ + ^ + 2/' = 0. 7. ^ + zrrz 2 y = 18xyl 



2. x^ + 3y + xy = 0. 8. ^ + yttmx = y*smx. 

dx dx 

3 - ~T + i — ~? V + 6 ^ = °- 9. -^ + xy = y 2 sin z. 

ax 1 — a: 2 ax 

4. «(*«■ + 1) d6 = 8da. 10 ^ (x¥ = L 

5. vr^l + ^^3. ^ - 

Solve by the substitution U " *a^ = ^S^ + ^ 

y = kc, and also by Set y — e v . 

x = sin a, y = sin /3. t _ , dy . , 

.-, 12. cot x -^ + ?/ log ?/ = rra/. 

6. coU^ + 2r = 4V~r. a+ f* 

d0 Set y = e v . 

293. Clairaut's Equation. Clairaut's equation is 

y = px+f(p), 

dv 
where p stands for -p, and /(p) is any function whatever of p. 

This equation is presented here because of the ease with which 
it is solved and because of the interesting form of the complete 
primitive. 

We first differentiate the equation as to x and get 

J r! 

where f'(p) = -r-f(p)- This equation reduces to 

[x+/'(p)]|=0. 

Neglecting the factor x-\-f r (p), which contains no derivative* 
of p, we have 

■j- — 0, whence p = c, 

* This factor leads to what is termed a singular solution. For a complete 
discussion of the very interesting subject of singular solutions, as well as for 
a full treatment of differential equations of the first order but of higher degree, 
see Chapters IV and V of Cohen's Differential Equations. 



§293 EQUATIONS OF THE FIRST ORDER 437 

and on substituting this value of p in the given equation we have 

y = cx+f(c), 

and this is the complete primitive of Clairaut's equation. Note 
that no integration was employed in the solution, that the com- 
plete primitive may be gotten by writing c for p in the given dif- 
ferential equation, and that it represents a system of straight lines. 



CHAPTER XXXV 

THE LINEAR DIFFERENTIAL EQUATION OF THE nth 
ORDER WITH CONSTANT COEFFICIENTS 

294. Definition. A differential equation of the nth. order and 
of the first degree in the dependent variable and all its derivatives 
is termed linear. Its general form is 

d n y d n ~ l y dhj dy _ 

Xo d^ +Xl d^+ ' * ' +Xn - 2 d^ +Xn - l d~x +Xny - X > 

where the X's may be constants or functions of x but do not con- 
tain y. 

We shall consider only the case where the X's in the first mem- 
ber are constants. Denoting these constant coefficients by /c's, we 
write the equation 
, . v 7 d n y . 7 d n ~ l y . , 7 d 2 y , , dy , 7 T;r 

We shall first consider the case where X = 0. 

295. The Solution of (A) when the Second Member is Zero. 

d T y 
For convenience we shall use the notation y^ r) for -=-^, and then 

when X = our equation takes the form 

(B) ; k yV> + kiy(»-V + . . . + h n - 2 y" + fe w _^' + k n y = 0. 

To solve this we assume y = e mx . Then 

y = e ™x } y r = me mx , y" = m 2 e mx , . . . ?/( n ) = m n e mx . 

Substituting these values in (B) and dividing by e mx , we get, 

(a) k m n + km" 1 - 1 + . . . + k n - 2 m 2 + k n - x m + k n = 0. 

This is known as the auxiliary equation. It is the condition that m 
must satisfy in order that e mx shall be a solution of (B) . But there 
are n values of m that satisfy (a), viz., the n roots of (a). Denote 

438 



§295 THE LINEAR EQUATION 439 

these roots by m h m 2 , . . . m n ; then e m * x , e m > x , . . . e m « x all satisfy 
(B) , and are particular integrals of (B) . Now if e miX is a solution, it 
can be shown by actual substitution that de miX is also a solution, 
whatever value the constant Ci may have. Further, it can be 
shown also, by direct substitution, that the sum of any number of 
such solutions as e miX , each multiplied by an arbitrary constant, is 
also a solution of (B). Hence having obtained n particular inte- 
grals by solving the auxiliary equation, a solution involving n arbi- 
trary constants is 

(b) y = Cie miX + c 2 e 7 " 2Z + ....+ c n -ie m ^ x + c n e™ nX . 

Therefore, by the principle of Art. 275, this is the complete primi- 
tive of (B). Hence to solve (B) we have only to find the roots 
of the auxiliary equation (a) and substitute them in formula (b). 
The difficulty in this process is of course the algebraic one of find- 
ing the roots of the auxiliary equation. 

d r ii 
Observe that the auxiliary equation (a) is (B) with -^ or y^ r) 

replaced by m r , and y by ra° = 1. 

Example 1. y" f - 2 y" - y' + 2 y = 0. 
The auxiliary equation may be written down at once and is 
m 3 - 2 m 2 - m + 2 = 0. 

The roots of this are 1, —1, 2, and consequently the complete solution of 
the given equation is 

y =r- c x e x + c 2 e~ x + c 3 e 2x . 

Let the student prove by direct substitution that this value of y satisfies 
the equation. 

The auxiliary equation is 

5 m 3 + 3 m 2 - m + 1 = 0. 

1 ± 2i 
The roots are —1, — - — , and therefore the complete primitive is 
5 

1+2 i 1-2 i 

y = c 1 e~ x + c 2 e 5 + c z e 5 , 

x f 2% x -2t \ 

= c x e~ x -\- e 5 \c 2 e 5 + c 3 e 5 J' 
Let the student prove that this value satisfies the given equation. 



440 DIFFERENTIAL EQUATIONS §§296-297 

296. Exercises. 

L &y_dy_ 20y==0 ^ 5. y" ±4^ = 0. 



dx 2 dx 



6. y'" ±4?/ = 0. 



2. 18 y" -27y'+ 10 y = 0. 7. </'" - 7y' + Qy = 0. 

dx 2 dx 9. ^iv — 5 2/" + 4 2/ = 0. 

4. 2/" ± 4 ?/ = 0. 10. i/ v - 2/' = 0. 

297. Complex Roots. When the complex number a + bi is 
a root of the auxiliary equation, the conjugate complex, a — bi, 
is also a root, because the complex roots of an algebraic equation 
with real coefficients occur in pairs (Art. 256) . Then two terms of 
the solution are 

(a) Cie (a+bi)x _|_ C2e (a-bi)x _. e ax( Cie bix _|_ c^-M*). 

Now by Art. 251, formulas (A) and (A'), 

e bix = cos bx -{- i sin bx, and e~ bix = cos bx — i sin foe. 
Hence (a) becomes 

e ax [(ci + c 2 ) cos foe + i(ci— c 2 ) sin 6x] = e ax (A cos fo + B sin foe). 
This can be put into still another form. We have 

A cos bx + B sin bx = K (-^ cos foe + ^ sin 6z )> 

where K = VA 2 + B 2 . Now ^?' ^ are a pair of numbers, the 

sunr of whose squares is 1, and these numbers are therefore the 
sine and cosine respectively of some angle a, and the quantity 
within the last bracket may be written 

sin a cos bx + cos a sin bx = sin (bx + a) . 

A B 

Or, the fractions ■= > -= are respectively the cosine and sine of some 

angle which we will call — /3, and the quantity within the brackets 
may be written 

cos jS cos bx — sin j8 sin bx = cos (bx + /3). 



§§298-299 THE LINEAR EQUATION 441 

We have then four forms for that part of the complete primitive 
which corresponds to the pair of conjugate roots a ± bi, viz., 
e ax( Cie bix _|_ C2e -bix^ e ax(A cos bx + B sin bx), 
Ke ax sin (bx + a), i£e ax cos (bx + /3) . 
The arbitrary constants are respectively, c h c 2 ; A, B; K, a] K, j8. 
Example. 5 y'" + 3 y" - y' + y = 0. 

This is the same as example 2 of Art. 295, where we got the solution in 
the form 

de~ x + e 5 \c 2 e 5 + c 3 e 5 / . 
We see now that this may also be written in the three following ways: 

X X 

Cie~ x -\- e* (A cos f x + B sin § .t), c x e~ x + 2Te 5 sin (f x -\-a), 

X 

Cie- x -\-Ke* cos (§ x + 0). 

298. Exercises. 

1. y" -2y' + 2y = 0. 4. */" - 4^ + 7?/ = 0. 

2. 2/"'-2/ ,, + 2/ / -2/ = 0. 5. yV + 5y" + 6y = 0. 

3. ^- V = 0. 6. ?/"-2?/" + 2?/' = 0. 

299. Multiple Roots. Thus far it has been tacitly assumed 

that the roots of the auxiliary equation were all different. We 

have now to consider the case of multiple roots. Take first an 

example, 

y"' -3y' + 2y = 0. 

The auxiliary equation is 

m 3 - 3 m + 2 = 0, 

whose roots are 1, 1, —2. 1 is a double root. 

From the argument of Art. 295, the complete primitive should be 

C\e x + c 2 e x + c 3 e~ 2x , 
which can be written 

(ci + c 2 )e x + c 3 e~ 2x . 

This is indeed a solution, but it is not the complete solution be- 
cause it contains only two arbitrary constants, Ci + £2 counting 
as a single constant. 

And so in general, if r roots of the auxiliary equation are equal, 
the solution obtained by the method of Art. 295 will contain a 



442 DIFFERENTIAL EQUATIONS §299 

term of the form (ci + c 2 + . . . + c r )e ax , a being the r-fold 
root. The sum c x + c^ + . ■ ■ + c r counts as a single arbitrary 
constant, and so the solution thus gotten will contain fewer than n 
arbitrary constants, and will therefore not be the complete primi- 
tive. 

Now the facts are these: If a is an r-fold root, then the corre- 
sponding part of the complete primitive is 

e? x (ci + c 2 x + c 3 x 2 + . . . + c r x r ~ l ). 

Thus in the above example, where the auxiliary equation has the 
double root 1, and the single root —2, the complete solution is 

e x (ci + c 2 x) + c 3 <r 2 *. 
To prove this general principle we proceed as follows:* 

Let the differential equation and its auxiliary equation be 
denoted respectively by E = and f(m) = 0, so that 

E = k yW+lc 1 y< n -Q+ . . . + k n - iy ' + k n y = 0, 
f(m) = k m n + kim/ 1 - 1 + . . . + k n -im + k n = 0. 

Suppose a to be an r-fold root of f(m) = 0. We now seek to 
determine u as a function of x so that e ax u shall be a solution of 
E = 0. Setting y = e? x u and differentiating successively, we have 

y = e? x u, 

y' = e? x [au + u'], 

y" = (f*[a 2 U + 2 au' + u"], 
y>" = (*x[ a 3 u _|_ 3 a V _|_ 3 au " _|_ w ///] } 

2,iv = e"*^ + 4 aV + 6 aV + 4 cm'" + w 1 *]. 

We see that the quantities within the square brackets are analo- 
gous to positive integral powers of a + u, and we are therefore able 
to write down y^> at once: 

n(n - 1) __ 9 ,, 






y {n) _ gax a n u _j_ na n-\ u > 



2! 



+ n(n-l)(n-2) at ^ v//+ _ 
o! 

+ n(n-l)...(n-r+l) rf „ _ _ _ +m(0) " 

r! 



See Cohen's Differential Equations, page 93. 



§299 THE LINEAR EQUATION 443 

Multiplying these equations, beginning at the bottom, by k , 
k h . . . k n and adding, there results 

E=r* [uf(a) + u'f'W + ^"^r + w/,,/ -^r + ' * ' 

r\ n\ J 

Now since a is an r-fold root of f(m) = 0, it follows from Art. 234 

that 

/(«) - 0, /'(«) = 0, . . . /M(a) = 0, /«(«) * 0. 

Therefore 

r! (r + 1)! n! 

and this is the condition that u must fulfill in order that e^u 
shall be a solution of E = 0. This condition is fulfilled if w (r) = 0, 
because then also u {r+1) = 0, u {r+2) = 0, . . . w (n) = 0. Hence 
u (r) = is the sufficient condition that e ax u be a solution of E = 0. 
But w (r) = is all that is necessary to completely determine u; 
for by successive integration we have 

u fr) = o, ul'-V =f a, w< r -9 = ax + b, u {r ~V = \ ax 2 + bx + c, . . . . 

Integrating r times, and using finally c's for the arbitrary coeffi- 
cients of powers of x, we have 

U = C\ + c 2 x + c 3 £ 2 + . . . + c r x r ~ l . 

Therefore that part of the complete primitive of E = that corre- 
sponds to the r-fold root a is 

ef* x (c 1 + c 2 x + c z x 2 + . . . + c r x r ~ l ). Q. E. D. 

Of course if there are other multiple roots, there will be other terms 
in the complete primitive similar to that just given. 

Example 1. y™ - 2 y'" + 2 y' - y = 0. 
The auxiliary equation is 

m 4 - 2 m 3 + 2 m - 1 = 0, 

which has a triple root, 1, and a simple root, —1. Therefore the complete 

primitive is 

y = e x (ci + c*c + c 3 z 2 ) + c^ - *. 



444 DIFFERENTIAL EQUATIONS §§300-301 

The principle just proved holds true if the multiple root is 
complex. Suppose that a + bi is an r-fold root: then the conju- 
gate, a — hi, is also an r-fold root, and the corresponding part of 
the complete solution is 

= e? x [cie hix -\-c 1 'e- hix +x(c 1 e hix +c< i .'e- bix )+ . . . +x T - l (cre hix -\-c r 'e- bix )\ 
= e ax [AiCosbx -\- Bismbx -\- x(A 2 cosbx-{- B 2 smbx) -{- . . . 

+ xr-t-fAr cos 6z + Z? r sin &#)] 
= e aa; (Ai + A 2 z + . . . + A^ -1 ) cos bx + 

e°*(J3i + J3 2 z + ■ • • + B r x r ~ l ) smbx. 

Example 2. ^ - 4 ?/'" + 14 y" - 20 ?/' + 25 ?/ = 0. 
The auxiliary equation is 

m 4 - 4 m 3 + 14 m 2 - 20 m + 25 = (w 2 - 2 w + 5) 2 . 

The roots are 1 + 2 i } 1 + 2 1, 1 — 2 f , 1 — 2 i. Hence the complete 

primitive is 

y = e x (Ai + A 2 x) cos 2 x + e x (Z?i + J9 2 z) sin 2 x. 

300. Exercises. 

1. y"'-3y" + 3y'-y = 0. 5. yv+5y"' = 0. 

2. yiv - 3 y'" + 3 «/" - y' = 0. 6. «r + 18 ^ v 4. 81 y" = 0. 

3. y Y -2 y" - 2 2/"'+ 7. ?/ IV - 8 2/'" -j- 24 y" - 

±y" + y' - 2y = 0. 32 y' + 16 y = 0. 

4. y* - 2 yiv + 2y"' - 8. ?/( 8 ) - 8?/( 4 ) + 16 = 0. 

4 2 /" + 2 / , -2?/ = 0. 

301. The Solution of (A) when the Second Member is not 
Zero. We shall write the equation in the form 

(A) hy in) + kiy {n ~ l) + • • . + K-iy" + K-iy' + k n y = X, 

where X is a constant or a function of ic, but does not contain y. 
We write again equation 

(B) hy {n) + W n ~ l) + • • • fc*-*v" + K-xy' + fcrf = 0. 

Let u be the complete solution of (B) ; u is termed the complemen- 
tary function of (A). Let v be any particular integral of (A) . Then 

hu^ + hu^-V + . . . + h n - 2 u" + fcn-iw' + fc nW = 0, 
fcbv<»> + hv( n -V + . . . + h n - 2 v" + fcn-iw' + k n v = X. 



§301 THE LINEAR EQUATION 445 

Adding, 

h(uM + »«) + h(u< n -V + v( n -V) +.-..+ kn-tiu' + v') 
+ /c n (u + v) = X, 

and this tells us that u + v is a solution of (A) . Moreover, since 
w + v contains the requisite number, n, of arbitrary constants, 
it is the complete primitive of (A). Expressing our results in 
words : 

The complete primitive of (A) is the sum of the complementary func- 
tion of (A), i.e., the complete primitive of (B), and any particular 
integral of (A). 

We have already learned that the complementary function can 
always be determined, and it is also true that the particular inte- 
gral can always be determined whatever the form of X may be. 
There are, in fact, several methods for finding the particular inte- 
gral,* but we shall explain only one of them, the method of un- 
determined coefficients, which is effective when X is of the form 

x 8 , e kx , cos Ix, sin mx, 

or when X is a sum of such terms, or a sum of products of such 
terms. The exponent, s, is a positive integer, and k, I, m are any 
constants. 

In such a case we write down the separate terms of X, and 
all the new terms that arise from successive differentiation of 
the terms of X, multiply each of these terms by an unde- 
termined constant, and assume that v, the particular integral 
sought, is equal to the sum of these products. We then take 
this value of v and its several derivatives and substitute them 
in (A) . By equating the coefficients of corresponding terms on 
each side of the resulting equation, we get a series of equations 
for determining the undetermined multipliers in our assumed 
value of v. These being determined, the form of v is completely 
determined. 

* See Cohen's Differential Equations, Sections 46-52. In particular see 
Section 50 for the method of undetermined coefficients employed in the text. 
This method is due to Dr. Cohen. 



446 DIFFERENTIAL EQUATIONS §301 

Example 1. y" — 4 y = 3 x 2 + cos 2x . 

The auxiliary equation is m 2 — 4 = 0, and its roots are ± 2. There- 
fore the complementary function is 

c,e 2X + c 2 e- 2X . 
Now the terms that occur in X are 

x 2 and cos 2 x, 

and the only new terms that arise from the successive differentiation of 
these are 

x, constant, sin 2 x. 

Therefore we assume v to be of the form 

v = ax 2 + bx + c + h cos 2 x + k sin 2 3. 
Then t/ = 2 ax + 6 - 2 ft sin 2 z + 2 & cos 2 z, 

0" = 2 a — 4 ft cos 2 z — 4 & sin 2 z. 

Substituting in the given equation, we have 

2 a — 4 ft cos 2 a; — 4 k sin 2 a; — 4 ax 2 — 4 fa — 4 c 
— 4 ft cos 2 x — 4 & sin 2 z = 3 x 2 + cos 2 x, 
or 

-4 ax 2 - 4 fa + (2 a - 4 c) - 8 ft cos 2 z - 8 k sin 2 x = 3 x 2 + cos 2 x. 
Equating coefficients of like terms in x, we have 

-4 a = 3, -4 6 = 0, 2a-4c = 0, -8 ft =1, -8 k = 0, 

whence 

a = — f, 6 = 0, c=— I, ft=— |, £ = 0. 

Therefore 

w= _|a;2_|_ i C os2:r =-K6£ 2 + 3 + cos2x). 

(The student should substitute this value of v in the given equation 
and show that the equation is actually satisfied.) 
Hence the complete primitive is 

y = Cl e 2x + c 2 e~ 2x - | (6 x 2 + 3 + cos 2 x). 

This method fails if X contains a term that is also a term of the 
complementary function. Thus if X contains a term e ax , sin a#, 
or cos ax which occurs also in the complementary function, a 
slight variation of the method is necessary to get the particular 
integral. This variation consists in using in the assumed value of 
v a term in xe? x , x sin ax, or x cos ax, instead of e ax , sin ax, t^os ax. 
And if a is an r-fold root of the auxiliary equation, we use x r e xx , 
x r sin ax, or x r cos ax. 



§302 THE LINEAR EQUATION 447 

Example 2. y" - 2 y' + y = x 2 + 2 e x . 

The auxiliary equation is m 2 — 2 m + 1 = (m — l) 2 and 1 is a double 
root. Therefore the complementary function is 

(ci + c 2 x)e x . 

The terms of X are x 2 and e x , and we see that the latter occurs in the com- 
plementary function and arises from a double root of the auxiliary equa- 
tion. Therefore we take as a term of the particular integral x 2 e x . More- 
over, it, is not necessary to consider the derivatives of this term, xe x and 
e x , because they occur in the complementary function. Hence we set 

v = ax 2 + bx + c + hx 2 e x . 
Then v' = 2 ax + 6 + hx 2 e x + 2 hxe x , 

v" = 2a + hx 2 e x + 4 hxe x + 2 fof, 
and 

v" - 2v' + t> = ax 2 + (6 - 4:a)x + c - 2 6 + 2 a + 2 fte* = z 2 + 2 e*. 

Equating coefficients of like terms in #, 

o=l, 6-4a = 0, c-26 + 2a = 0, 2h = 2, 
whence 

o=l, 6 = 4, c = 6, h = 1, 
and 

v = z 2 + 4x + 6 + z2 e * # 

The complete primitive of the given equation is then 

y = ( Cl + c 2 .t> x + z 2 + 4x + 6 + z 2 e*. 

This method fails also in case X contains a term of the form 
x s e ax , x s sin ax, or x s cos a£, where a is an r-fold root of the com- 
plementary function and s is a positive integer. In that case the 
corresponding term in the assumed value of v must be of the form 

X s+r e ax } x s+r g j n a% ^ Qr %8+r cog aX 

302. Exercises. 

1. y"-y' + y = yx 2 . 8. y" - 4 y = 3 x 2 + e 2X . 

2. y"'+y"+y'+y=a?+2x-22. 9. y" + 4:y = x + cos2x. 

3. 2/" + 3 y' + 2/ = f cos x. 10. ?/" - 4 j/' + 4 = mxe x sin a;. 

4. 2/'"+ #' + 22/ = a; sin a;. 11. y" -2y' = mxeTsmx. 

5. 2/ IV -2/ = 45sin2x. 12. y" - 5y' + 4y = 4:X 2 e 2x 

6. 2/" - 2 y' + !/ = -4 e x sin 2 a;. + 10 x sin 2 a;. 

7. y" -2y' + §y = e x $m2x. 13. 2/"'- 3?/ +2?/ =4 a; 2 (9e x +l). 



448 DIFFERENTIAL EQUATIONS §303 

303. Two Special Equations. The two following equations are 
of rather frequent occurrence and are readily solved: 

(I) -p- = X, X being a constant or a function of x only. 

By successive integration we have 

c[n-iy r* d n ~ 2 y C C 

■^Ti=JXdx + C 1} fa^=J JXdx-dx + dx + Ct, 

13 = fff x ** + cix2 + C2X + cs > 

etc., etc., etc. 

d 2 s 

(II) -p = S, where £ is constant or a function of s alone. 

ds 
Multiplying this equation by 2 -j dt, we have 

Now the first member of this equation is identically d ( -rr J ■ There- 
fore 

Since £ is a function of s only, the last integral is also a function of 
s only. Call it <f>(s). We have then 

-£\ = 0(s) + c and ^ = Vtf>(s) + c, 

whence , = = dt, 

and on integrating again 

J V<f>(s) + c 
which is the complete solution. 



§§304-305 THE LINEAR EQUATION 449 

This equation can also be solved as follows: 

p ds _ ,, d 2 s _ dv _ tfo ds _ dy 
dt ' dt 2 dt ds dt ds 

and the given equation takes the form 

dv 
v -r~ = S or 2vdv = 2Sds, 



f*2vdv = A 



whence / 2vdv = I 2 S ds + c 



and 



v 2 = f2Sds + c = 0(s) + c, 



and we proceed as before. 

This equation was integrated by this method in Art. 151. 



304. Exercises. 










1. -7* — sm 2 x. 
dx 2 


4 -fH- 




7. 




ig-A*,,. 


, <Ps 

5 - » =g - 




8. 


dx*~ 6 ' 


3. T7 = COS 0. 

dd* 


Solve in two 
a d 2 s „ 

6 - * = as - 


ways. 


9. 


9 ^?/ 1 

cos 2 £-7-^ = 1. 
dx 2 



305. Miscellaneous Exercises. 

1. (2 x 2 — 2 xy + ?/) efc Q. y' — y = C os z — sin z. 

-x(l + x-y)dy = 0. 7. vT=^d a + \ / r r ^d i 3=0. 

2> _^/_4_J/^_ = . 8. 2/ VI — 6?/ IV + lly"— 6y = 0. 

X + y x ~ y 9. 2/"-3?/' + 4?/ = 6e 3a: cos2z. 

3. 2 /' + 2 2/ cotx = 2mcsc^. 1Q y „ _ 3y , + 4y = i 6a 4_ 4a a 

4. 2/ ' + 2 2 /cotz=-2mcsc2z. n ( , + g3) ds _ ( ^ _ fe2) ^ = Q 

5. (sin«y-scosy)dy 12 ( 2 ^ - %' + 1 + jf = 0. 

+ sin y dx = 0. 

13. (w Vw 2 - v 2 - y Vl - ^ 2 ) dw + u Vl -u 2 dv = 0. 

14. -M= + ( l - ^=U=0. 

V x 2 + y 2 \y y Vx 2 + y 2 / 



450 DIFFERENTIAL EQUATIONS §305 

15. x log x ~ + y = x. 

16. y" + 3y' + 3y = 13sinz. 

17. y "-y = 2e x . 
18> y ' J r 2ytsmx = yicsc 2 x. 23 - z" + 4z = sin2x + cos2z. 
19. y'"-Zy"+Zy'-y = 6e*. 24 - ^l-z 2 )y'-?/ = 3z 3 U-z 2 ). 

dy = y(x 2 y 2 + y 2 + 1) 25 - 2/'" - ?/" - 4 ?/ = 16 x cos 2 a;. 

cte z(xV — 2/ 2 + 1) 26. y" — 2 y = mxe x sin x. 



21. ^- S - 
eft 2 


ds 
- 2 — - + 3 s = wfe* sin Z, 


22. y'- 


, , ,y + 3xy 3 - Q. 





ANSWERS 


Art. 14, page 24. 




1. 0., 2. 1. 


3. 1. i. 0. 6. 1. 


9. 1 10. f. 


11. |(a + l). 12. 3< 



6. 1. 7. 0. 8. 0. 

13. fa. 14. fa 2 . 

Art. 24, page 37. 

1. 2x; yi + y = 2 Xjx, 2 x x (y - y Y ) + x - x x = 0; 

?/-4x+4=0, 4 ?/ + x - 18 = 0. 

2. 6x + 2;j/-8x + 4=0;?/ + 4x + 4=0. 

3. -i; 4*/ + x + 8=0; */ + x - 4 = 0. 

4. 2 ?j'?.f ;4y-5x + 3 = 0, 10y + 8x-13 = 0. 

(x + 1)^ 

5 f 6 (re 2 - x); z/ = 0, x = 0; y + 1 = 0, x - 1 = 0; y - Y2x - 7 = 0, 
12 y + re + 61=0; 4?/-18x + 27 = 0, 9?y + 2x -3 =0; 
2/ - 12 x + 20 = 0, 12 y + x - 50 = 0. 

Art. 27, page 43. 

1. 3x 2 + 2x + l. 

2. 6 re 2 -8rr + 1. 

3. x 3 - x 2 - 1. 

4. 12 x + 19. 

5. 6w(5m 3 -2u- 3)Z)u. 

6. 3 (2 w 2 - 3 u + 7) 2 (4 w - 3) Dw. 

7. 5(2x-3)(3x -2) 2 (6x -7). 

8. 6 x (2 x 3 - 3) (3 re 2 - 2) 2 (12 x 3 - 4 re - 9). 

9. 3(x + l)(x + 3) 2 (x-5) 3 (3x 2 -ll). 

10. u' = 6 (re- 3) (re - l) 2 (re 2 - 5). 

11. D<£ (v) =2v(v - 1) (y + 3) 2 (3 v + 5) Dy. 

12. j/' = 60 re 2 (re 3 - 4) 4 (5 x 7 - 12) 2 (3 x 7 - 7 x 4 - 3). 

Art. 29, page 44. 

1. */ = 2(x 1 + l)x-x 1 2 , 2(zi + l) (y-yx)+x-xi = 0' } y+l = 0, x+l = 0. 

2. y = 0, x = 0;2/-3x + 2=0, 3*/ + x-4 = 0; 

?/ - 12 x - 16 = 0, 12 y + x + 98 = 0. 

3. ?/ = 0, x =0; 2/ = 2(a;-l), 2y + x = l; y + 2(x + l)=0,2y-x = l. 

4. ?/i?/ = m (xi + x), m (y — y{) + y x (x - Xi) = 0. 

5. xix — yiy = 1, x x y + yix = 2 x x y x \ 2x=F V3y = 1, V3 x =b 2 j/ = 4 Vs. 

6. 3x=F2j/-2=0, 2x±3y — 10 = 0. 

7. 3y :: F2x-2 = 0, 2?/±3x-10=0. 

451 



452 DIFFERENTIAL AND INTEGRAL CALCULUS 

8. 3?/+x=f4 = 0, 3z-?/=F2 = 0. 

9. 2*/ + 3x=f8 = 0, 3?/-x=f1=0; y + 24x=F32 = 0, 24?/ - x=f191 = 0. 

10. ?/ + x-l =0, x-y-1 = 0; 4?/ + 1 = 0, x = 2; 

4?/-2x-7 = 0, 4?/ + 8x + 13=0. 

11. ?y (?/i 2 — ax x ) + x (xi 2 — ayO = ax x y h 

x (yi 2 — axi) - y (xi 2 - ay x ) = (2/1 - Xi) (xi?/i + axi + ay{). 

12. 3?/ + 5x--8 = 0, 5?/-3x-2 = 0; by + 3x + 8 = 0, 3?/-5x -2 = 0- 

Art. 32, page 46. 

. - x 2 - 2 x + 1 x 2 -2x-l „ x 2 - 6 x - 2 



(x 2 + l) 2 (x 2 + l) 2 (x 2 + 2) 

21 x 2 - - (4x 3 + 15x 2 +6) e 3x 2 

«>• 7-= ^r^ • O. 



(3x 3 + l) 4 (x 3 -3) 2 (x 3 - 3) 2 

- x (x 3 + 3 a; - 2) g 8a; a: 2 (3 - x 4 ) 

(x 3 + l) 2 ' ' (3-x 2 ) 2 ' * (x 4 + l) 2 ' 

10 1 ~ 4x " 11 ~x 2 (2^ + 3) -(x 3 -6x 2 + 2) . 

(x 5 + l) 2 ' (x 5 -l) 2 ' a: 3 (x + l) 2 

13 -Qc 3 + 2x 2 -6) -x6 + 6x 4 + l 2x(l-x) 

x 3 (x-3) 2 ' (x 4 + 3 x 2 - l) 2 ' (2x + l) 4 ' 

1ft -*(« + *) 17 * 2 (3 ~ x 2 ) -x 2 (3+x 2 ) 

1D * (x-2) 4 ' (x 2 + l) 3 ' (x 2 -l) 2 ' 

-2(x + DQr 3 + 3a, 2 -l) 6a; (3a; 2 + 1) (3 - a; 2 ) 

19t (2x 3 + l) 2 ' (x 2 + 2) 4 

12x(2x 2 + l) 2 (x 2 -3) 15 a; 2 (a; 5 - 2) 2 (x 2 + 2) 

21 * (3x 2 -2) 3 " (a; 3 + 1)6 

Art. 34, page 48. 

2 (2 Vx + 1) 2a; 3 - 4 m 2 Z)m . 

Vx" ' V2 x 2 + 1 ' (l-2u 3 )* 

- 4 (1 - 2 u 1 ) 2 Du e - Dy - (a 1 - x* 1 )* 

4. T 0. 7= , -• D. r • 

u h 4 Viy VI - Vy a; 3 



2/(1 + w)* vl - a; 2 a; 2 V x 2 - 1 

a. 7 , * 7 _ • 12. *J 

Vx (1 - Vx) 2 
uv — x {uv' + vu') , m/ + v — (u + x) v' 



10 . 2 » 2 ~ 3 n „ 11. _i — _. 12. »(~ / -y0 + «>, 

t; 4 VI - v 2 Vx (1 - v x) 2 v 2 



13. 



2„2 



try 



14 x<£' (x) - </> (x) . (x) - X0' (x) 



x 2 [0 (x)] 2 

i6. * . 17. — ^==. is. *:-»» ■ 

(l-x)Vl-x 2 (l-x 2 )Vl-x 4 2x*V0 

u (2 xtm' — xwv' — 3 mv) < u (2 xvu' — 3 xuv' — m>) 

t^x 4 ' v~W 

(2 MX 2 + It 2 - x) ?/- (2 xu 2 + x 2 - u) 
(x 2 + w 2 ) 2 



ANSWERS 453 

21 ^-\){vu' -uv')+u^-v^ 2(gf - fg') 

(vx + uy ' (/ + gY 

OA a?-aVa?-x* „ -2(l + Vl-x*) 
24. . . 25. — . 

x 2 V a 2 - a: 2 x 3 v 1 - x 4 

2g x(2u + x)u' +u(2x + u) 

3 (u 2 x + wx 2 )* 
Art. 36, page 49. 

1 rv i r> i ^ vD v u—u n _ 1 — 3 x 2 — 2 x 3 n 

1. vD x u-\-uD x v: v+uD u v; — ; u+xD x u. 2. — , , , iX9 — D u x. 

i/ 2 (z 3 + \y 

3. i3 /±l_ D,3,. 4. — 2* 6. V * + ^W 

2 y i Vy« - 1 (1 + Vx) 3 2 Vl + x 2 

- 3 + 2Vx „ SD e u Q 20 



6 Vx ( Vx + 1)* ' (3 - w 2 ) § ' (1 + 1 ) 4 

9 z (z - 2) i)^ 1Q 16^(32-1) 

( 2 2 + l) f (2z 2 + l)*' ' (9z 3 + 4(S)3z 2 + 4)s' 
Art. 38, page 50. 

_ ». + tfa- W - 2> _»(^' + 2) 3. -iori. 4. -2orl0. 

u 3 + wx 2 — x x (w 3 x + 2) 2 10 

5. 2/ + x-3a = 0, y-x = 0; 5y-4x-4a = 0, 12y + 15x -26a=0. 

6. y + x =f —7= = 0, y — x = 0; y = 0. x = ± a; x = 0, y =±a. 

V2 

7. ^~ X - 8. 4xi2/ + (xi - 3 y x ) x = xi (xi + yi). 

9. y=Fa = 0; 5 V3y = 3x + 8a. 10. y = 0; 2/TV2 = 0; x = VI. 
Art. 40, page 53. 

1. 6 tan 2 2 x sec 2 2 x. 2. 1 =F 2 sin 2 x. 3. 2 sin 2 0. 

4 cos Vu p m B cos yDjy , sin yD^y 

2Vm ' 2 Vl + sin y ' 2 Vl - cos y 

6. tan 2 z; —cot 2 z. 7. 2 cos 5 x — s in 2 x sin 3 x. 8. cos 3 ?/. 

9. sec 4 y. 10. 2 Vsec 2 2 tan 2 2 ; - 27 2 esc 3 3 3 cot 3 3 . 

11. xcosx; xsinx. 12. (csca — 2 csc 3 a) Dta. 13. cos2uD x u. 

1A DyU ._ DyU «- . _. 

14. -. 15. , , v . 16. tan 4 wDw. 

l-smw 1 + cos u 

17. sin 5 x. 18. tan 5 x sec x. 
Art. 42, page 55. 

1. 2 2 . DtU 3 DxV 



Vl-4x 2 V2w-w 2 1+2/ 

a 1 2 * 2 

4. 5. -■ 6. 



2 x Vx - 1 (x + 4) Vx (x + 4) Vx 



7. i/«±»- 8. 2VT=> 9- 

V a — x 



(t + 1) V* 2 + 2 * 2 



454 DIFFERENTIAL AND INTEGRAL CALCULUS 



10. 11. j- • 12. 



xiVS^Ti ' (1+z 2 ) 2 (1+x 2 )' 

13. -". 14. l 15 



(x + l)Vx (z+2)V2x + 3 1 + xi 

Art. 46, page 58. 

2. 1/ xv= 24(5-*). = 2 (8 fl»- 18 *' + !). 

4. D 4 Vx = - — ^-= ' 5. D n xs = s (s - 1) . . . - n + 1) x*~ n . 

16 x 2 

9 * 18 (i+^)^ ' 13 * yyYl + 6 ^ V + 15 y " ylV + 10 {y " y " °" 

15. wvi = (6 u 2 - 1) i*iv + 48 ut*V" + 72 u'V + 36 u«" 2 . 

Art. 53, page 68. 
1. 3.176. 2. 2.858. 3. 2.116. 

4. ± (75° 52' 14"), ± (265° 53' 29"). 5. 50° 13' 58". 

6. - .641. 7. 63° 50', 158° 51'. 8. in each case. 9. .399. 

Art. 61, page 78. 
1. Max., 8, min., 4. 2. Max., 1, min., | and — 15£. 3. Neither. 

4/b + 1 

4. Sin x has its max. value, 1, when x — — ^ — -k, and its min. value, — 1, 

4/c + 3 ,' 
when x = — ~ — x, k being any integer. 

5. Max., 1, min., — 1. 6. Max., — 1. 

7. Max., - (a + 6) 2 , min., - (a - 6) 2 . 

8. Max., 7= > min., 9. Max., V3 v^ mm -> _ V3 -^2 

2V6-6 2V6 + 6 

10. Max., 2 V3 </% and 0; min., - 2 V3 </2, and 0. 

Art. 63, page 80. 
1. Max., 38, - 16, min., - 38, 16. 2. Min., .637. 

3. ^Min., ~ ^P 2 ■ 4 - Max -> ^t, min., 0. 

6. / (fcrr) =(-.!)* is min.; / (± | + 2 far) = - is max. 7. Neither. 

8. When* =£ +for, « «* (- l)*^andisS maX, jwhenfcisS e 7; n j. 

4 4 (mm. ) (odd ) 

9. /(|+ kir\=2(- l)*-lismin.;/MJ| + 2farj = | is max.; 

/ f -g- + 2 /ctt J = 2 is max. 

13. / (0) =5 is min.; / (1) = 7 is max.; / (2) = - 11 is min. 

14. / (0) =7 is max.; / (± 1) = 6 is min. 



ANSWERS 455 

Art. 67, page 84. 

1 a - 2 - - 

l " 2' 2 2' 2 

7. Radius = -j» arc = =• 8. Convex surface = -~— 

9. Sum of alt. and base radius of cylinder = \ alt. of cone. 

4 ttt^ 

10. (1) Volume = =; (2) convex surface = 2-n-r 2 . 

3 v o 

11. In both cases alt. of cone = f r. 12. In both cases alt. = diam. 

13. Alt. = radius of base. 14. Alt. = —p. • 

V3 

15. (1) Area = 144; (2) perimeter = 48. 

16. (1) Volume = 486 tt; (2) convex surfa ce = 14 4 it. 

17. (1) Area = 2ab; (2) perimeter = 4 V a 2 + b 2 . 

18. Radius of cylinder = (1), ~\ (2), — 

V6 V2 

19. (1) Length of intercepted portion = a + b; (2) area of triangle = ab. 

20. Length of ladder = 21.2 ft. 21. Length = {al + &f)f . 

22. Radius of tent base = 4 V6; height = 4 V§. 

23. Angle of sector cut out = 66° 3' 45". 24. 12 ft. 26. -^-> 

V2 

26. Distance from center of sphere of radius r is — = ~ * 

Ri+rl 

27. A hemisphere. 

30. Greatest profit $5.40 at a depth of 36 ft. 

31. 9 miles per hr. 32. 18 miles per hr. 
33. 21.55 mi. per hr. 34. 15.44 mi. per hr. 

35. 1st, 2 hrs. after noon; 2d, 2 hrs. before noon; 3d, 24 min. before noon; 
24 min. after noon. 

Art. 71, page 94. 

1. 2{x + l)a* 2 + 2I loga; 2 (x + l)e x2 + 2X . 

2. loga eu • D x u; (1 + log u) D x u = log eu • D x u. 

3. av(l+y\oga)D x y; eV{\ + y)D x y. 4. 3*(log* 2 + l); tet>(2 + St*). 

i 

5. ~— 2 Du; - 6x5- 3 * 2 log 5. 6. cot0; -cot0. 

7. - tan 0; 2 esc 2 0. 8. - — - Du; -\og2-Du. 

2Vu 2v« 

9. asin u cos u \ g a J) u - e t&nu sec 2 u Du. 
10. (1 — u sin u) e<» s u D U} ( cos u _ s i n w ) e w j) Ut 

U . 2_^. 12 . 1 _. u. ± D " 



3 (** - 1) (1 - x)Vx Vtf^~l 

U *££.. 16. ^a.. 16.29-?- 

e 2 * - 1 e* + 1 



1:56 


DIFFERENTIAL 


AND 


INTEGRAL 


CALCULUS 


17 


1 




18. 


1 




19. (1 -tan0) 2 . 




2Vx 2 + 2x 


1 + V x 


20. 


1 




21. 


tan -1 u • Du. 




22. 9x 2 logz. 


xVx + 1 


23. 


Min.,|(l-log|y 




24. 


Min.,-i. 




25. Min., -A. 


26. 


Max., — 1, min., ± 1- 


-e ±l . 27. 


4 

Max.,-, min, 

e 2 


,0. 


28. Min.,-??. 

e 3 


29. 


tv/t 256 

Max., —7- , mm., 0. 




30. 


Min., e. 




31. Neither. 



32. Min., 2. 39. Curve has a discontinuity. 

Art. 78, page 107. 

1. a cos - t. 2. 112 ft., - 112 ft., 2500 ft., 400 ft., 25 sec. 

r 

3. 85 ft., 40 ft., 5 ft. 4. 0, - I - 5, - |> 0, |> 5, |> 0. 

6. 132 ft. per min. 6. 131.93 ft. per min. 

7. 26.18 ft. per hr., 52.36 ft. per hr. 

8. 20 miles per hr. 9. 11.33 miles per hr., 25.9 miles per hr. 
1Q 4(50* ±27) 

VlOO f> ± 108 1 + 81 

11. The velocity of projection upon a diameter is proportional to the sine 
of an angle. 

12. 45.42 ft. per sec; - 51.89 ft. per sec. 

13. 2| ft. per sec; 14 ft. per sec; if the top of the ladder were to con- 
tinue in contact with the wall (which it cannot do), the velocity would 
be 00 . 

14. 48,000 7T sq. mi. per hr.; 47,856.32 tt sq. mi. per hr. 

Arty83, page 112. 

1. 2t and 2-wr times as fast. 2. 871-r and 47rr 2 times as fast; when r = 2. 

3. v = a + 2bt + 3 ct 2 ; a = 2 b + 6 ct. 

4. v = — ka sin (b + kt) ; a = — k 2 a cos (b + kt) = — k 2 s. 

7. />.„ — *,D„ — £. 8. z>.p-i2-^A_. 

Art. 104, page 148. 

1. -• 2. — t - 3. 3v*W- 4. D 8 = 2aem$0= ^2^y; D x s =V~ * 
y x* T y 

6. y 6. 2 V£. 7. |. 8. 2(o + 6)«n^;2(a-5)aniLft 



ANSWERS 457 

Art. 107, page 152. 

1 /tr ; — „ /pr— o ds V2 ra/c Op rafc sin 



3. - times the point's distance from the ground 



dt (i - cos a)* dt c 1 - c ° s e y 

rom 

d^ _ — kay dy _ kbx 
dt~ b (a 2 - e 2 x 2 ) h ' dt ~ a ( a 2 _ e 2 x 2)* ' 

ac at at p 

Art. 112, page 161. 
1. r = a , a = = 0. 2. r = V% « = 0, /3 = If. 

3. r = jU*(4 + 9s) 1 , a = - | (9* + 2), = i*? (3 s + 1). 

4. r =-> a=x -yy' =x -V- Vy* - a 2 , /8 = 2 y. 

6. r = 2 V2, a = 3, |S = - 2. 6. r = 2 V2, a = - 2, = 3. 

7 . r = (^_^f; („)l + ^)l.(„)l, 

_ (e 2 x 2 — a 2 )* . a l * 

8. r = r — — » (az) 3 - (&2/) s = (ae) 3 . 



9. r = ^(l -e 2 cos 2 0)*;a =^cos 3 0, = -^sin 3 0. 

10. r = (x2 ^/ 2)g ; (x + y)l - (2 - y) 1 = 2 a*. 

11. See Arts. 109 and 110. 

a . 2 a cos 3 _ 2 a sin 3 

12. r = / > a = ~5 / > P = ^ — / 

3Vcos20 3 Vcos2 3 V C os20 

13. r = ad; x = a cos d, y = a sin 0; x 2 -\- y 2 = a 2 . 

14. a = a (0 + sin 0), (3 = — a (1 — cos 0). 

15. 3a = a (2 cos + cos 2 0), 3 = a (2 sin + sin 2 0). 

Art. 117, page 171. 
1. -9. 2. f. 3. oo. 4. -f. 5. 0. 7. f. 8. 0. 

9. §. 10. oo. 11. 1. 12. -• 13. 1. 14. 1. 15. log a. 

16. 2. 17. -• 18. 1. 19. 0. 20. 0. 21. (a), 0; (6), 0. 

7T 

9 2 a/9 9 

22. -• 23. 0. 24. 0. 25. -— -• 26. 0. 27. 0. 28. -. 

TV O 7T 

29. -1. 30. 1. 31. J. 32. 3 a 2 . 33. 1. 34. 1. 35. 1. 
36. 1. 37. -7T. 38. 1. 39. oo. 40. 1. 



458 DIFFERENTIAL AND INTEGRAL CALCULUS 

Art. 119, page 173. 

1. 1. 2. 1. 3. e. 4. 1. 5. 0. 6. 2. 7. log 2. 

8. e. 9. - I. 10. 1. 11. - 1. 12. 0. 13. e^. 14. |. 

16. log a. 16.0. 17. -• 18. 0. 19. ~ 20. -1. 21. 2. 

e Ve 
22. - 1. 23. No limit. 24. 0. 

hit. 125, page 183. 

1. x 4 - 2 x 3 + 2 x. 13. - 2 Vl -z. 23. u s - log (u 3 + 2) 2 . 

2 - ^L 14. |(« + 5) 4 . 24. f V> - 3. 

4 ^ ogic2 15 « t(- 4 -^-3) § . 25. MlogO 2 - 

5. log V^l. 16 ' * *' " t * 5 - i * 4 26. | tan 2 x. 

6. i(y*-l)K + fx 3 -x. 27. §sec 2 x. 

s ° 18. log (1 - 22/ + 1/ 3 ). 

a i„„^ , 1 i 29. 2 VI - C os0. 

8. logx+— • 19 1 

9. |[^+log(^-l)3. 1 - 2 2/ + 2/ 3 30 ' Hsin-^. 

10. Jlog(x 3 -l). 20. §(l-2 2 / + 2/ 3 ) 1 . 31. log V4 + 3 tan d. 

11. x l+l . 21. | (1 - 2 2/ + t)K 32. - cos 0. 

1 22. ix 5 -fx 4 + x 3 -x 2 

12. 



4 (x 2 + 5) 2 + x. 



'■iKJ 



Art. 128, page 186. 



1. itan-i?. 9 . sin- 1 ^- 19 ' 2 sin- 1 Vx or 



2— 2 



sin- 1 (2 x - 1). 



2. -sin-^x. 10 . Igec-i?^!. 20. log (V a + l) 2 . 
* 4 2 

oil, , x « OA Aj — - 2 21. log(x 2 + l) 

3. i-^tan-i- 11. -2V 4 -0 2 . -Stan^x. 



V3 V3 



12. tan-!e u . 



1 • 1^2 ^ ,— 22. sin- 



x — 2 

4. -^sin-i-^. i V21 *"' ollx " _ ^ _ ' 
V3 V2 13. — =tan- 1 -^-x. 

V 21 7 23> _V 4 x-x 2 . 

5. -V2jf + 3. 14> l log(3a .. ±7) . 24. tan-i(y + l). 

o 

6. stan-^. 1 „§ 25. tan- 1 ^- 1 
6 16. ^tan- 1 !-- 



1 . V6x 



26. 3 sin- 1 * 



7. -^sec- 1 -^- 16 . 2tan- 1 Vi. -' OBm ^"3" 

x z2 17. sec- 1 2 0. _ VfTx^ 

8. — tan- 1 -^. . 

2V5 V5 18. -2vl-M. 27. sin- 1 tan x. 



ANSWERS 459 

Art 130, page 187. 

x _a^_ 5 . TL^L. 9. i(e* + l) 4 . 

' log a*" log a 1Q log(e „ + 1) . 

2 -- 5 - 5 _ 6 -;;+r- ii. log (*.+«>. 

3 * ~ e ~" ; "ioiS* 7 ' loi^' 12. *-log(e« + l). 

4. e 8inx - 8. ie^ + fe^+S^+x. 

Art. 132, page 188. 

1. -±cos3z. 8. tantt-u. 13 . h og (a - 6 cos x) . 

2. jan gs-l ). 9 _ (cotw + w) . & 

3. logV S ec2.. 10> |tan:c 3. £ secw - 

4. fsin 4 0. _ , 7 15. -icsc4x. 

5 __j_. "' -21ogcosV^+l. 16 |seG 3 2M> 

2sin 2 ^ 12# itan(a + 6x); 17. Hz - sinz cosx). 

6. — 2 esc ^ • 1 18. \{x + sin z cos x). 

7. -J log cos </ 2 . - l cot(a + bx). 19 a cos 3x-cosz. 

Art. 133, page 188. 

1. V2x - 1. 22 - sec- 1 ^. 

. _ 2x-a 23. - | log cos (y* - 1). 

A Sm a ' 24. -| cos (^_l). 

3. ilogsin(a + mz). 26 * lo S tan *- 



m 



a L 



4. \ e *v + e™ + y. 26. 

6 - 6(l-n)(t + te)- 1 ' 27 ' 1 °S V 2^M- 

R f .6£ + l 28. -|(1_V5)1 

°" tan 2 * 29. I (log sin w) 2 . 



7. f (sin t)* . • 30. sin- 1 x - Vl - x \ 

8. !x 3 + z 2 + 4z + log(x-2)9. 1 2x 

N 1 x 31. |log(2a: 2 + 3)-^:tan- 1 ^- 

9. ix 2 -log(x 2 +2) + -^tan- 1 -^- * V6 V6- 

10. Cannot be integrated by the 32. -f V3-2x 2 — sin- 1 -^- 

foregoing methods. v 2 V 6 

11. ilog(x. + 9)+|tan-| 2 . » ~ tt'^M 2 ^° r l^ 

o 34. x 2 + 2 x + | log (4 x 2 — 4 x + 5) 

12. |x + |sin(3x + 2)cos(3x + 2). +ta n- 1 (x-|). 

13. tan -1 sin x. , . x — 2 

14. log(e*> + </ 2 ). *• -3V4z-z 2 + 4sin-i-2— 

15. - f cos 3 z. . x - 3 

16. cosz -fco s 3 z. 36 * sm ~2 — 

« -/^-^m-Ml-2 W ). 37# _ V 6x-x 2 -5 + 3sin-^. 

19. log(x + 2) 2 . 38. -V3a;-a: 2 -2+fsin- 1 (2a;-3). 

20. T 9 o (2 s - 3) (re + l) 1 . 39. 3 sin" 1 0\ 

21. 2sin- 1 ^- V^u-u 2 . 40. Vo" tan- 1 3 X ~t 2 • 

2 V5 



460 DIFFERENTIAL AND INTEGRAL CALCULUS 

Art. 135, page 192. 

i. loga-zHj-^. ii. -^(i-^+^(i-^ 

1 . V7+a?V3 1 ~ . 4 

3 - 2^ l0g vr^vr 12 - -^^SA- 

6. -i=log(V3a?-7 + xV3). 14 . _ J__ (8 + 9 „■) (4 _ 3 M i } « . 

1,1, 2 + x 1 1 

6- ~4^ + i6 log 2^ r x* 15 « g sin 5 cos - — sin 3 cos 



8. ^xV3-4x« + i sin- v _ 



2 1 . „ „ , 1 

2z 



7. vz2 — 4 _ 2 cos- 1 -- — — sin0 cos0 + -»d. 



ifi 1 sin0 1 /x , 0\ 

16 ' 2c-0^ + 2 l0gtan U + 2j' 

9 * 2LrT^ +tan ~ l2/ J' 17 ' |tan*0-itan 2 0-logcos0. 

3(l+w 3 ) ^lS ° g (l+w) 3 18. ^(4x-36 2 )(2o: + 6 2 )^ / 62 x _ x 2 

1 V3 , 2m-! 6 6 . ,2a; - 6 2 

H — 7t- tan x 7= — 4- — sm — 1 

9 V3 T 16 6 2 

Art. 138, page 194. 

1. * [a* -log (a* + 1)]. 9 4 ( Vs + 2) 

2. 2 [v^ - log (l + Vy)]. "" a/Vs + 1 " 

3. T 2 7 (3a;-2)(.x + l)l 10. f log [fo + 5) 1 - 1]. 

4. 2tan~ 1 JF_ 11# ^ (4 Vi - 3) (Vi + l)* 
. _ 1 Vfl + 1-1 / -7= 

5 -> g V0+T+ V 12, - 4Vl ~ V ^ 

6. A &z-3Kx+ l) f . 13 ' See Art " 135 ' ex - 18 ' 

2 (x + 2 ) 14. - T 2 , (3^ + 4^ + 8) VT^x. 

V^T+I s a + 4 

8. f(x-3)(:r + l)*. ' V2 + x 2 

Art. 140, page 196. 

\ 4 cos 2 

1. I cos 3 x — cos x. 4. — — log cos 0. 

2. - |sin 3 x +sinz. 5. ^- (3 cos 2 s - 7) cos* 2. 

3. \ sin 5 - § sin 3 + sin 0. 6. § (cos 2 2 - 5) V 



COS 2. 



ANSWERS 



461 



Art. 142, page 197. 



1. ^(sin- 1 * - zVl -z 2 ). 



Vl -4x 2 



'J 



+ sec 



-) 



4. 



2 x 3 - 3 a 2 x 



3 a 4 (x 2 - a 2 ) 2 

2 a 3 |_w 2 + a 2 aJ 

6. V3 2 2 - 2 - V2 sec- 1 V} 2. 

7. See ans. to Ex. 15, Art. 135. 



Art. 

1. 

2. 

3. 

4. 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 

24. 



144, page 200. 

xsinx + cosx. 

(2 — x 2 ) cos x + 2 a; sin x. 

x tan x + log cos x — \x 2 . 

\{x — sin x cos as) . 

|(x + sinx cos re). 

e x (x 3 -3x 2 + 6x-6). 

x tan- 1 x — I log (1 + x 2 ). 

(x + 1) tan- 1 Vx - Vx. 

| [3a; 3 sin- 1 a: + (2+a: 2 )Vl-x 2 ]. 

-e-*(x 2 + 2x + 2). 

u log u — u. 

\ e z (sin z — cos 2) . 

- T %(2 + 3z)(l-z)\ 



14. 
15. 
16. 
17. 
18. 

19. 



I (2 1 sin 2 * + cos t sin * - *). 
\ sin 3 cos — \ sin cos + 1 
Ax 2 (logx 2 -1). 

-§(6+t0(4-»)» 

a; 6 



h log 



7 + 5a; 6 



.-iE±2. 



log(x-2 + Vx 2 -4x + 12). 

i + vr+^ 



21. Vl + 7/ 2 -log 



2/ 



22. 
23. 



z(2 + z) 



2 log (1+2). 



1+2 

3*5 (5 sin x — sin 5 x) . 



l ri. (2 + w) 3 /5 + i^-n 

i2L2 los ^+^ +V3tan ^rJ 



ri. 8 + ^ 3 

L2 l0g (2+^ 



in 

6 



25. 
26. 1( 
27. 



+ V3 tan- 



V3 

wj-1 

V3 



] 



Vx-1 



2x 



+ tan- 1 Vx - 1. 



28. 
31. 



8(l-x 2 ) 2_r 
Vl+*x 



I log 



1 + x 
Vl+x-1 
Vl+x+1 



30. |(4- sin 2 6)8111*0. 



: h (35 cos 3 0-42 cos 5 + 15 cos 7 0) , 

V4 + 5x-2 



32. flog 

V4 + 5x + 2 

33. - ^ (5 x 3 + 12 x 2 + 32 x + 128) vT 

34. f (x* + l)° (3x f -2). 

35. T s iJ (3 + 2xV(x l -l). 



2x. 



462 DIFFERENTIAL AND INTEGRAL CALCULUS 



X2 

ke 2 . 


a 

5. y = he mx 1 




8. xy = k. 



36 . L+2z ^_ log( ^ + 1 + Vl+2x-z 2 ) + logx-sin- 1 ^i 

x y/2 

37. i T2 rr V25-4z 2 + 25 sin- 1 ^1 • 

38. J [2 x V4 x 2 - 25 - 25 log (2 a + VIx^ 25)]. 

Art. 146, page 204. 

1. y = ke ax . 2. y = kx a . 3. ?/ 2 — ax 2 = e. 4. 
6. A system of parabolas. 7. xy = c; x 2 — y 2 = 
9. The logarithmic spiral. 

10. 1st the circle, 2d the parabola, 3d the equilateral hyperbola. 

11. p 2 = a 2 sin 2 d. 12. The cardioid. 

Art. 148, page 208. 

2. 140 ft. per sec; 506 1 ft. above the ground; 180 ft. per sec. 

3. 1st, 6^ sec; 260 ft. per sec. 2d, 10 sec, 260 ft. per sec; rises to a 

height of 1056 j ft. above the ground. 

4. x = a{l -e^)- 5. -^— = atf. 

a — x 

6- "-f^jf^aV; ^U=(»-l)«»-^ + l. 7. 178.12 ft. 

Art. 150, page 212. 

1. 4.92 miles per sec; 42 min. 32 sec. 4. 54 min. 18 sec; 1.04 miles per sec. 

5. If s = when Z = 0, s = ^- (e* - e"^) and v = 5 ( eM< + e~ M 0- 

Art. 152, page 215. 

2. y, is too great by about (a) 1|%, (6) ^%, (c) 2<nny%- 

3. yist, 4.92 miles per sec; 34 min. 48 sec. 

2d, 6.82 miles per sec; 33 hrs. 7 min. 22 sec. 

4. 1.47 miles per sec. 5. 1.4 miles per sec; 11 hrs. 32 min. 9 sec. 

Art. 154, page 220. 
1. 120 ft. per sec, 112| ft., 5.3 sec. 2. 441.7 ft. per sec, 762.1 ft., 13.8 sec. 
3. 1299.85 ft. per sec, 2\ miles, 57.4 sec. 4. 2828.4 ft. per sec, .176 sec. 

5. 1.4 miles per sec, 52' 5", 7.07 sec. 

6. (a) 324.4 ft. per sec. (&) 75.6 ft. per sec. 

7. 9° 20' and 80° 40'. 8. 124.9. 9. Range = v \/—' 15. 5.78 miles. 



ANSWERS 463 

Art. 156, page 223. 

1. About 9 ft. from the ground. About 84°. 

2. v = v e-w. s = ^(l-e-* 2 <). 

3. If v = and s = when t = 0, 

. v = _J! Vl-e-2^. A; 2 s = log (e* v ^ + e~* v ^) - log 2. 

1 , 

* = ^7= log (efc«« + V e 2^ s _ i). 



6. -1. 
12. | log 2. 



Art. 161, page 230. 
1. 25f. 2. 19i. 3. 3f . 4. 6|. 5. log 4. 6. 3.206; 2.38. 

7. 5f. 8. | a 2 . 9. io a 2 . 11. 2; 2 V2. 12. Area of arch = 6f. 

13. Areas of the several regions are 1£, 4.7, 8.1. 

14. -Vx. 15. 4tt. 16. fV3andfV3- 



Art. 158, page 225. 








1. 3f. 2. -3|. 


3. 


0. 4. f. 


5. 1. 


z 3 

7. S 3- 8. -5|. 


9. 


I" 10 ' 2 ' 


u 64 V2 
15 


13. log 2. 14. ^- 


15. 


7T 

32* 





Art 
1. 


. 163, 
53f. 


page 233. 
2. If 


3. 3tf. 


4. 5*. 5. 3 Tra 2 . 

4 


6- |. 


7. 


fa 2 . 


8. fa 2 . 


9. fa*. 


a 2 / - --\ 
10. o"(e a - e «)• 


11. Tra 2 . 


12. 

Art 


fTra 2 
.166, 


13. |xa6. 
page 235. 


14. 73f 


IB. i*. 

8a 




1. 


xa 2 
12' 


2. Tra 2 . 


Tra 2 

3 ' "8"' 


4. f (tt-2); a 2 . 


_ 5 Tra 2 
5 * -32- 


6. 


67ra 2 


7. a 2 . 


*f>- 


-3V3), «*<»£* 





Art. 168, page 238. 

2. 4V3- 3. 6 a. 5. 8 a. 6. 16 a. 7. 2Tr 2 a. 8 20.54 a. 

9. a. 10. ^[2V3-log(2-V3)]. 11. ^ [V2 + log(V2 +l)]. 

13. 8.17 tt. 14. 3.74 tt. 



464 DIFFERENTIAL AND INTEGRAL CALCULUS 

Art. 177, page 260. 

1. f 7TT 3 , 4 Trr 2 . 

2. ^ (j8 - a) (3 r 2 - a 2 - /3 2 - /fa), and 2 tit (/3-a), where x = a. and z = /3, 
o 

are the equations of the planes. 

3. Ua\ ^2 + ^logi^y 

4. 1st, \M\ 2d, $xa&«. 5. |. 6. ^ 

7. T VWa 3 , V^ 2 . 9- i^a 3 , 127ra 2 . 

10. 7 = 2 x 2 r 2 A, S = 4 7r 2 r/i, where ft is the distance from the center of the 

circle to the axis of revolution; V ' = 2ttV 3 , $ = 4-7r 2 r 2 . 

11. For upper semicircle, V = tit 3 (-\°- + tt), S = 2 Trr 2 (tt + 2). 
For lower semicircle, F = ttt 3 (-^°- — ir), S = 2 Trr 2 (ir — 2). 

12. 7=47ra[(6-a) 2 + fa 2 ], S = 4Tr 2 a(6-a). 

13. (1) x (2 - V2) a 2 . (2)7rV2a 2 . 15. ^-? Tr 2 a 3 , 6 V^Tra 2 . 

o 

16. (1) ^==- (2) 2 7ra 3 . 17. -^—TPa\iPa\ 18. 57r 2 a 3 , - 6 /- xo 2 . 
V2 3 V3 

19. 6 7r 3 a 3 , 16 7r 2 a 2 . 20. Tr 2 a 3 , - 3 g 2 - zra 2 . 

21. f tt(9tt 2 -16), f7r(37r-4)a 2 . 22. 2*0*. 

23. W, 2x[V2 + log(V2 + l)]. 24. 3tt 2 . 

25. 2 (2 n - 1) 7T 2 , 2 ttV. 

26. About OX, ff 7r and J| 7r; about the double tangent, 3 V3 tt and f V3 tt. 

27. shiraV, T 8 7 7ra 2 6. 

28. The number that gives the volume is ~ times the number that gives 

A 

the surface. 

29. (1) WaK (2) 00. 30. ^a\ ^-wa\ 

31. i^a 3 , T \ Tra 3 . 32. (1) J 7r 2 a 3 . (2) \% iraK 

33. ^7r 2 a 3 . 34. (1) j^tt. (2) ff x. 

Art. 179, page 264. 

1. - 6 f xa 3 , if ^ Tra 2 . 2. tt (2 - V2) a 2 , x V2a 2 . 



3. <ln both cases, y^xa 3 . 4. T § 7 (8 V2 — 9) xa 3 . 

6. 32V ^ „,3, 6 . A ^3 J^L [i 4 V3 + log (2 - V3)]. 

105 6 v 3 

Art. 184, page 270. 

1. (4,-2, -*),(*, -3, -|),(-i, -1,|). 

2. (o, - -v-, - f), (- H, 0, tt), (-1,-2, 0). 

3. Vertices. Edges. 

a = (2, 1, 2) ab = VII be = V26. 

= (3, 0, -1) oc = 3 bd = V24. 

c = (- 1, 1, 2) ad = VII cd = Vli. 
d = (1, -2,3). 



ANSWERS 465 

4. x-y -52 + 3=0. 6. 3x + 4y + 2z - 5 = 0. 

7. 11 2/ - 5 z + 2 = 0, 11 x + 14 2 - 21 = 0, 5 x + 14 2/ - 7 = 0. 

8. No. 10. ziz + y# = r 2 , ftjy + ** = r 2 . 11. (f, 0, |). 

Art. 191, page 283. 
1. AA> + BB'+CC' = 0;^,=^, = ^,. 

2. ' L = VI = * ; /A + mB + rcC = 0. 
A £> L 

3. x = Xi + Ar, y = y x + £r, z = z x + Cr. 

A 2 +£ 2 + C 2 ' A 2 +£ 2 +C 2 

A(Az x -Cx x )+B(Bz x -Cy x )-CD 
A 2 +B 2 +C 2 

4. Z (x - xi) + m (2/ - yi) + n (2 - z x ) = 0. 5. x + y + 2 z = 4. 

6. (mitt 2 - ^2^1) (a; - x x ) + (niZ 2 - n 2 Zi) (2/ - yd + (l x m 2 — hm x ) (z — z x ) = 0. 

7. I, f, f; x = 1 ^ r, y = - 1 + 2r, z = 1 + 2r, 

or a; = 3 + r, 2/ = 3 + 2 r, 2 = 5 + 2 r; = sin- 1 —^« 

V6 

8. -JL> -JL, -jL- 9. 2a; + 5y + 3a =0. 

V38 V38 V38 

10. x + y ± V6 2 = 0. 

11. Perpendicular, z = ar + Zr, 2/ = y + mr, 2 = 2 + w 

where I, m, n, satisfy the condition, 
I (xi - x 2 ) + m (yi - 2/2) + n (z x - a 2 ) = 0. 
Parallel, 2 = x + (a* - a; 2 ) r, y = y + (2/1 - 2/2) r, 2 = a + (z x - a 2 ) r. 

12. = cos-if|. 13. No. 

Art. 193, page 289. 

1. f 7ra6c. 2. 7ra 2 6c. 3. f mr 3 . 4. irmr 3 . 5. -V 6-7-3 - 

6. %irb 2 a. 7. fTra 3 . 8. 87m 3 . 9. j\ira 3 . 10. fTrafcc. 

Art. 195, page 291. 
1. 4r 2 . 2. 2 7ra 3 . 3. ^-^a 2 . 4. 8a 2 . 5. 2 a 2 (71- -2). 

6. fa 3 . 7. ^-a 2 . 8. -J [2 V3 + log (2 + V§)] a». 

9. V2 + log (1 + V2). 10. |(x+2). 

Art. 200, page 299. 

1. xix + yiy + 212 = a 2 ; jb = a* (1 + r), 2/ = 2/1 (1+ r), 2 = a x (1 + r). 

3. z + z x =k {y x x + Z12/); a; = x x + fc^r, 2/ = 2/1 + &W, 2 = z x - r. 



466 DIFFERENTIAL AND INTEGRAL CALCULUS 

4. xi 2 x + yi 2 y + Zi 2 z = a 3 ; x =xi (1 + x x r), y = yi (1 + 2/if), z = Zi (1 .+ zir) . 

5. xi 2 x — y{-y = a 3 ; x = Xi + 2i 2 r, y = yi — yi 2 r, z = z x . 

6 - i + ^ + 1 =0; x =Xi+ h r ~ v = 2/i+ i r ' z = Zi + h r - 

7. ZiZ = yix + Xi^; x = Xi + ^r, y = yi + Xir, z = z x - Zir. 

8. z — = xi — 2; a; = Xi + r, ?/ = y x r, z = z x r. 

2/i zi Vi zi 

9. — i + A + —r = a}) x = xi + — r, y = 2/1 + — j r, z = z 2 + — r. 
^i s 2/1* zi* X! 3 y^ z x * 

10. At the points where Xi = 0, and 2/1 = any integer, 

z = 2/iz; a; = 2/ir, 2/ = 2/i, z = — r. 
At the points where Xi = ~ , and 2/1 = 2 m (an even integer), 

z = (2 ma; + ^ 2/ - 2 m?) cos m^; 

x = -+2m cos mw -r, y = 2 m + - cos mwr, z = —r. 

At the points where £1 = -> and 2/1 = 2 m + 1 (an odd integer), 

A 

. 2 m + 1 7T _ lt . 2 ra + 1 
z = sin 5 — *j a; = 2, 2/ = 2 m + 1, z = sm w — r. 

At the points where X\ = ir, and 2/1 = any integer, 

z = (yix +iry — 2 ynr) cosync', 

x = ir + 2/1 cosz/iTr-r, 2/ = 2/i + 7rcos2/i7r-r, z =— r. 

11. At the point (0, 0, 0), 

x + y — z = 0; x = r, y = r, z = — r. 

At the point { - > ~ > J , 

/ + z =tt ; 
^t the point f |> — |i J 



a; + 2/ + z=7r; x=-+r, y=^-{-r, z = r. 



x + y - z = 0; x = |+ r, a; = - | + r , z = - r. 
At the point [ J> ^ > 1 ) > 

2 = 1; z = -, ?, = -, z = 1 + r . 



At the point (j, — p j, 



a; + ?/ - z = 0; x =^ + r, y = - 1 + r, s = - r. 

At the points (tt, ±x, 0), 

£ + ?/ — z=7r±7r; x=7r + r, 2/=±7r + r, z= — r. 



ANSWERS 467 

Art. 202, page 301. 

d 3 U d 3 U 

3. — =-ycos(x + z), — = -xcos(y + z), 

d3u / . ^ ( . ^ 

— = - x cos (y + 2) - y cos (x + z), 

a s w . , . . d 3 u . , , . 

- sin (x + 2), — — - = - sin (y + 2), 



dx 2 dy v ' " a^a?/ 2 

-^- 2 = - sin (y + z) - y cos (x + 2), 

— — - = - x cos (y + 2) - sin (x + 2), 
o?/d2 J 

d 3 U 

sin (?/ + 2) — sin (x + 2), 



dxd?/d2 
9 3 u 
x 2 az 
2xz d 2 y 2xz d 2 y x- 



d 3 u , . , d 3 U 

_«.-.y COB (* + *), —=- XCOB (y + Z ). 



dX 2 (x 2 + 2 2 ) 2 0Z 2 (x 2 + 2 2 ) 2 dxdZ (X 2 + 2 2 ) 2 

a 2 2 _y 2 - r 2 a 2 2 x 2 - r 2 d 2 2 xy 



ax 2 2 3 dy 2 2 3 5x3?/ 2 3 

5 2 2 d 2 2 

6. ^- 2 = y% + «)(1 '+ e«») e^, ^ = x% + z)(l + e^) 6*1/, 

d 2 2 

^- = (?/ + 2)(1 + x?/ + xye^) e*». 

Art. 204, page 306. 

dz cos (x - y) 1 + log < 

*' dt l + t 2 °* ' 1 + log*)*' 

d2 _ /5/ 5w , 5/3y\ dx ,/5/ du 3/ ay\ dy 
d£ ~ \aw ax aw ax/ d£ \,aw dy dv dy) dt ' 

dx ax a?/ dx a2 dx 



7. 



10. 



11. 



a/ a^_a/ <fy d l 9i _ d l d l 

dy _ dx dz dz dx dz_ _ dy dx dx dy 

dx ~ df dg _ df_ dg' dx ~ df dg _ df_ dg' 

dz dy dy dz dz dy dy dz 

dy _ dx dz dx dz _ dy dx dx dy 

d* E^T~I^^ , dx~ ^4.^/^ 

dy + dzdy dy dzdy 

dz_ _ dy dx dy 
dx~ dj_d$ _dfd$' 
dy dz dz dy 



X 2 



dy 


dz^ 


dfd$ 
dx dz 


dx 


dfd4> 


dfd<f> 


dy _ 
dx — 


dydz 
y(z-x) 

X 2 


dzdy 

dz 

' dx" 



468 DIFFERENTIAL AND INTEGRAL CALCULUS 

Art. 208, page 313. 

1. Let ux + vy = A, uy — vx = B, u 2 + ^ = M, x 2 + y 2 = N. 
. v du _ A to _ _M, du _N dx = _A 
W dv ~ B' dv ~ B' dy~ B' dy~ B' 

C ; dv ~ A' dv A ' dx ~ A ' dx ~ A' 

^ du~ A' du~ A' dy~ A' dy~ A' 
, ,, ft) _ _ i dy _M. dv_ __N dy_A 
W du~ B' du~ B' dx~ B' dx~~ B' 

^ e) du~ N' du~ N' dv~ N' dv~ N' 



(d) 



dx ~ [dx^ dv dx' dy ~\dy)' du~\du) dvdu 



W dt ~ \ du dx + dv dx) dt ~^~\dudy + dv dy) dt 

dz = 4yr + 1 dz _ y + s dz = dz = 

dr 4yz ' ds 2yz ' dr ' ds 

7 ( \ du _ x — v 2 dv _ vy — u # du _ ux — v dv _ y — 
dx uv — xy' dx~ uv — xy' dy uv — xy ' dy 



4yr 


+ 1 


4 

x - 


yz 

-v 2 


uv - 

y - 


-xy 

u 2 


uv - 
uv - 


-1' 
- 1 



r 


uv — 


xy 


X 


-V 2 




UV 


- 1 






uv 


— xy 



,,. dx _ y — u 2 dy _u — vy # dx _ v — ux dy _ 

du UV — 1 ' du UV — 1 ' dv UV — 1 ' dV 

. . dy _ uv — 1 dy _ x — v 2 ^ dv _ _y — u 2 dy 

dx v — ux' dx ~ v — ux' du v — ux' du v — ux 

dx dv ' dx du ' dy dv ' ' dy du ' ' 
where Ds f/|2_2fL. 

du dV dV dU 



9. 



d 2 u _ d 2 v _ 2 (xy - uv) . d 2 M_ d 2 y _ 2Q + a- 1) - u 2 - v 2 
dx 2 ~ dx 2 {u - v) 3 ' dy 2 ~ dy 2 ~ (u - v) 3 

d 2 u d 2 v xy— x — y 



I dxdy dxdy (u — v) 3 

y 

Art. 217, page 331. 

sin a 5 2 

Art. 226, page 346. 



1. 2S = a 2 {ir-2). 2.^-. 3. ^<z 2 . 4. ^^to 2 . 5. f (3 V3 -7r)a 2 . 



1. x = y=z = \r. 2. x = %r, y = z = 0. 3. i = f a, y = z — 0. 

4. x= T Vaj = z = 0. 5. £ = § a, y = § 6, z = f c. 6. x = o~,y = — 



3 



7. x = § a, y = f V2 ma. 

8. x = f a, y = z = 0. 9. (1) x = .51; (2) x = .636 I 

10. The point of intersection of the medians. 

11. The point of intersection of the altitudes. 



ANSWERS 469 

, _ _ 256 a „ _ 21 _ _ _ .. _ _ . 

12. x = y = 3j^;- 13. x = j28 a > y = z = °* 14. x = 7ra, y = f a. 

16. 5 = Tra, y = f a. 16. x = — f a, iy = 0. 17. x = — | a, ?/ = 0. 

18. x = — fa, V = z = 0- 19* x = .555 a, ?/ = 0. 20. ^distance from base. 

__ _ 2 r sin a _ «*> - - - 2 oo - r» - ^ + ac 

21. x = 3a , y = 0. 22. x = ?/ = z = | r. 23. x = 0, y = -^- • 

«>. - n - 2 & 2 + /ca + a 2 OK - - - a 
24.s = 0,y=3 fc + a . 25. *-,,-,-- 

26. x=z=^,y=0. 27. x = y = .27 a, z = .86 a. 

_ 128 a _ 4V3a 

105 7T O IT 

Art. 233, page 361. 
1. See Ex. 4, Art. 231. 2. See Ex. 6, Art. 231. 

i.^-» + * , + aI+-8f+-.;-+-sr + * 

e ex n — 0) n e0* 

where B = ^y^ or J-^*— *•+•• 

4. See Ex. 2, Art. 229. 

1 2 1 2"-isin(n-l)| 

5. sin 2 x = x 2 - j* +-x« - — x* + . . . + - { x»+R, 

2 n sin (n? + 2 9x\ 2 n (1 - 0) n sin (n^ + 2 dx\ 

where R = / ,,.. ^x»+i, or p ^x"+ 1 . 

(n + 1)! n! 

o 1 2»-i S in(ra-l)| 

6. eos 2 x = l-x 2 +3X 4 --xe+ 3 -^x 8 - . . . - } x»-R, 

where R has the same value as in Ex. 5. 

8. f(a+h) =c Q +c 1 a+c 2 a 2 +c 3 a z +(c 1 +2 c 2 a+3 c&*)h+(c 2 +3 c z a)h 2 +c s h 3 . 

Art. 239, page 374. 

1. Vl -x 2 = 1 --x* --x< -- 4 x* - — ,xs - _^-__ai. _ 

3.5-7.. -2n-3 

2»-n! -h • • • • 

2. When x 2 < 1, 

,/- • 1 1 . 2 , 2-5 B 2.5-8 . 2.5.8-11 ln 

1 * 3 3 2 -2! 3 3 -3! 3 4 4! 3 5 -5! 

2.5.8. . . (3n-4) 

3»-n! X -r- . . . . 

When x 2 > 1, 

V1 * * ^ W V i + 3x 2 + 3 2 .2!x 4 + 3*.3!x6 + 3M!x* + 

2-5.8 ... (3 n- 4) \ 

•••"*■ 3 n -7i!x 2n + •••;• 



470 DIFFERENTIAL AND INTEGRAL CALCULUS 

Art. 242, page 378. 

1. Divergent for all values of x. 

2. 3. 4. Convergent for j x | < 1, divergent for [ x | > 1. 
6. Convergent for all values of x. 

6. Convergent for — 1 < x = 1. The ratio test shows the convergence 
for — 1 < x< 1, and the convergence for x = 1 was shown in Ex. 2, 
Art. 238. 

7. 8. Convergent for all values of x. 

Art. 244, page 379. 

x 2 . 2x 4 . 2 4 z<5 , 2 4 .17z 8 , 28.31x 10 , 

1. logsecx = ^ + Tf + -e r + — gy— + 1Q! + 

_ , . . 2x 2 , 2x 3 2 2 x 5 2 3 x 6 2 3 x 7 , 2*x* , 2 5 x™ , 

2. ^3;=* + — +_-_-_- _ + _ + w + .... 

- , x 2 . 5x 4 . 61 x 6 , 1385x8 , 

3. se cx = l+2] + Tr + -g r + -g^+ 

A sin* 1 1 , X * 3 * 4 8x 5 3x6 

k cosx fi z 2 ,4x 4 31x6 379x8 \ 

5. ecosx = ^i__ + ____ r + ____ ...j. 

ci /i i • n x 2 . x 3 2x 4 . 5x 5 

6. log(l+sinx)=x-2 ] + 3 ] -- ir + - ir - 

„ , , X 3 , X 5 X 7 , X 9 

7. tan-'z = z-3+ y - T +g- .... 

Convergent for | x | < 1; divergent for | x | > 1. 
y. sin x-x + 3! i- 5 , -t- ?| -f g , -f u| -f . . . . 

Art. 246, page 382. 

3. Natural logarithms. Common logarithms, 
log 5 = 1.609 437 912 log 5= .698 970 001 
log 7 = 1.945 910 148 log 7 = .845 098 035 
log 11 = 2.397 895 272 log 11 = 1.041 392 680 
log 13 = 2.564 949 356 log 13 = 1.113 943 346 
log 17 = 2.833 213 342 log 17 = 1.230 448 915 
log 19 = 2.944 438 977 log 19 = 1.278 753 594 

Art. 248, page 386. 

1. e («+*)(«H-*) = e x * y °\l+hk + \ h 2 k 2 +A + h A 2 

+ ±A'+^A* + hkA + hhkA*] 

+ ^ [60 h 2 k 2 + 20 hkB 2 + £ 4 ] e(*°+* A ) («»+**) 
where A =hy + kx , B = hy + kx Q -\-26hk. 



ANSWERS 471 

3. e*°+ A sin (y + k) = e*° [~s + (hs + fee) + h (h 2 s -\-2hkc- kh) 

_J_1 (/ l 3 s + 3/ i 2^ c _3^2 s _/ c 3 c ) + ^(/ i 4 s + 4/ i 3 /cc _ 6 / l 2/ c 2 s _4/ i / e 3 c+fc 4 s )J 

+ ij(/i 5 -10/i 3 /c 2 +5Afc 4 )sin(^o+^) + (fc 5 -10/c 3 /i 2 +5/c/i 4 )cos(^o+^)]e^^, 
5! 
where s = sin ?/ , c = cos y . 

6. -Omitting primes, 

/ (xi + /ii, x 2 + h 2 , x 3 + /i 3 ) = 

/ (xi, x 2 , x 3 ) + 2 [/ii (an^i + ai 2 x 2 + ai 3 x 3 ) + ^2 (a 2 iXi + a 22 x 2 + a 23 x 3 ) 
+ h 3 (a n xi + a 32 x 2 + 033^3)] +f(hi, h 2 , h z ). 

Art. 262, page 392. 

1. sin 4 0=4 sin cos (1 - 2 cos 2 0), „- _ _ . ±VS + i 

cos40 = l-8sin 2 0cos 2 0; * - *, ■ 2 

sin 5 = sin 0(16 sin 4 0-20 sin 2 0+5), . ± V3 - ; 

cos50 = cos 0(16 cos 4 0-20 cos 2 0+5). V-i= + i, 



\/2 + V2 - i V2 - V2 \/2 - V2 + i \/2 + V2 

3. ±- 2 ' =*= 2~3 

4. -2, 2 (cos 36° ±i sin 36°). 5. ±2, 1 ± i V3, -l±iV3. 
6. ± V5(. 4472+. 8944 i). 

Art. 268, page 402. 

1. log(x + 3) 3 (x-2). 4. ±x*-2x + ±\og^' 



(a + l)« 



n . (x-3) 7 K . x 3 (x-3) 

2. log)- ^f- 5. log 



(x-2) 6 "* iW& (x + 3) 2 

fr + 2) 
(a; + 1) (*-«' 



3. }log(x-4) 4 (x + 3) 3 . 6 . Ilog ti^wf,^ - 



Art. 260, page 403. 

1. l 0g( ,-3) 2 -^3. 2 . log^_ + ^. 3.1og^±l-l 

4. log(z-l)3- 7 -^— • 7. log^- + 



(x-1) 2 " 8 1 -x ' 2x(l -x) 

x 2 , 4x + 3 3x + l 

6 « l0 § ^ 1 1 >2 + O ^ I 1 \2 • 8 « ~ 



(x + 1) 2 ' 2(x + l) 2 " 6(x + l) 3 

„ .. x-1 x _. 1 x . i , x -\- 

6. ilog— r-r-07^2— TV 9- <r^ irT-T 2 + tt. 1o S 



x + 1 2 (x 2 - 1) 2 a 2 x 2 + a 2 ' 4 a 3 ° x - io 

Art. 262, page 405. 
1. logx + tan^x. 2. log-^=-i. 3. ilog^±i + l tan-^. 

1. (x-1) 2 1 . ,2x + l _ 1, x 2 -x + l 

4. 6^fc+i-vl tai1 ^T' 7 * 2 log x^+7+r 



472 DIFFERENTIAL AND INTEGRAL CALCULUS 

x 2 + 1 x 

5 - 1 °g a . 2 _ 1 _ 2 ' 8 " 2 tan_1 2 ~ tan_1 X ' 

« 1r ,„^+l , 3 , 1 x 0+ . 1, l+«.l, 

6 - l0g x^F2 + vl tan vi" 2tan " la: - 9 - i log r^ + 2 tan x - 

10. -log 1 log — 

4 x 2 + x -1 4 z 2 +V5x + l 

Art. 264, page 408. 

z 2 + 2 V2 V2 

z + 2 •. x 2 + 4 1 . . .a; 

2 - -4(^+4) +l0g -^-8 tan 2' 
3. Slog ^ 2(3^ + 1) A 1+2** 



V^+I ( X -I)(a? + 1) 4(x 2 + l) 2 

_ . re . 1 2z 2 + 3 

5. log - + 



V^+i ' 4(x 2 + l) 2 
e . x 2 . 9x 3 + 8z 2 + 15x + 4 . 9 + . 

6 - l0g ^+i + — s^+w — +8 tan - 1 - 

Art. 266, page 409. 

1. \ ["log * + 1 + V3 tan- ^il • 
3L Vx 2 - re + 1 V3 J 

2.5- log ^ + v 3 tan- 1 -— • 

3L Vx 2 -x + l V3 J 

3. \x* - hog(x 3 + l). 4. hog(2a;-3)(6:r 2 +z-15) 2 . 

OO O 

- 1 , x + 1 . 7 , x - V6 „ , a* + z + 1 1 

- lj ,1-1 „ 1, x<(* + 3) 2 „ If. a: 2 , o "1 
12. ^log (x 2 + 4x + 5) - 5 tan-' (s + 2) + \ ^ 2 + X ~ x ' + g ■ 

13 - F7I L log VxT+i + tan r^J • 

u . i r log feta: + 2 v 3 - tan-' ^ + 4^] . 

9 L x 3 + 1 V3 z 3 + U 

Art. 268, page 410. 
1. y = lx+VlW + b\ 2. | + *=1. 3. ,-|5* + |. 



Art. 271, page 415. 

1. y =±x. 

2. y* = ± 3 V3 x 2 . 

3. y 2 = 2 mx. 

4. x 2 + y 2 = a 2 . 

5. y 2 = 3 s ±1. 

6. ,y = x + l. 

7. y = ev*-K 16. p = 2 a V^nT0. 

8 * y = x - 16. p = 4aVir~20. 

9 - » = «loga;- 17. P = asm2e. 

10. x?/ = ^- 18. p=r(l+cos0). 



11. 


x 2 + y 2 — mx = 0. 


12. 


?/ 2 =2mx + ra 2 . 


13. 




14. 


P = — ■» sin tan 0. 



473 



20. ?/ 2 = 2m(m-i). 



2 
Art. 273, page 418. 

11. x = | [1 + e 2 + (2 - e 2 ) sin 2 0] cos0, y = 6 sin 3 0. 

12. ?/ 4 -6:n/ + 3 = 0. 

13. a; = | (3 cos 2 - e 2 ) sec 3 0, y =-b tan 3 0. 

14. z = | (1 -cos 2 0), ?/ = | (20 -sin 20). 

Art. 277, page 425. Art. 282, page 428. 

1. xy" —y'=0. 1. X y = c; y = ex 

2. y-xtf + vr^j*. 2 - x + y = ™y- 

3. x 2 y" - 2 xy' + 2 y = 2 x 3 . 3. x = cev 

4. y" + y = 



4. cos cos 



5 «"-« = () 5 * lo S c ^+ f — 7= = °- 



6. z/'" -Gy" + lly' -Qy = 0. 6. sec y = ce GOS x . 

7. e~ x - e~ y = c. 



8. u- C + v 



7. (1 + y' 2 Y = ry 

8. (l+y">)y"' -Sy'y" 2 =0. " " ~ 1 - cv 

9. xV' + x*/' -*/ = (). 9 * ™ y = *ye x - 
10.xy>» + 3y»=0. 10 ' (log^ = log C -^ 
Art. 284, page 429. 

1. x (xy - 2) = cy 

2. x 2 + - = c. 






3. x=cey-x. 8 Vx 2 -\-y 2 tan-^ = c. 

4, 3,-atan-i^= c . -1= 

a 9. x = ce ^ - * • 

x V 1 + x 2 + VV + y 2 = ex, 10. x 2 + y 2 =?xcQse + ysmc. 



474 DIFFERENTIAL AND INTEGRAL CALCULUS 

Art. 286, page 431. 

1. xy = sin -1 3 + c. y 

2.x(xy-2)=cy. 8. s* + ** + 2 taa- f 

3. y = xta.nl - x +cY 9 - ^ + tan-i| = c. 

10. 4> sin + 6 sin = c<j>9. 



A 9 I X 

4. a: 2 + - = c 



» 



Art, 


, 288, page 433. 




V 


1. 


y = cex. 




-J) 


2. 


y = ce x . 


3. 


(?/ — x) 2 = X 2 + c. 


4. 


log (?/ 2 - rn/ + x 2 ) = 


5. 


-—tan- 1 y 
V3 V3z 
z 3 (y — z) = c. 


6. 


W 2 

y 2 = ce ^ 2 . 


7. 


log z = sin -1 - + c. 



4. |A = (1 - X 2 ) (x + c) 



c. 



2/ 11. 2/ sin?/ + 1 



5 - x2+ y = C - 12. V^^+ S i n -i * = c . 

6. sin- 1 - + wy = c. 13. sin" 1 ?/ + - = c. 



y 



7. az 2 + zV = cz 2 . 14. y + x esc w = c. 



8. zcos - = c. 
z 

j/ 

9. a; + ?/ = cez . 

10. y = xe* 

X 

+ c. 11. (x — y)e*-v = c. 

i 2 .i og5 = ^li!) + c. 

14. x 4 — y* = c. 

Art. 290, page 434. 

1. xy = sin- 1 x + c. 10. ysin 2 x=x + c. 

2. z 3 (?/ - x) = c. 11. e S ec = tan - + c. 

3. (2/ - z 2 + 1) Vl - z 2 = c. i2. ( w _ 1) gsin 



c. 

13. ?/ = 1 + cesin- x z. 

5. p = y 3 (y + c). ^ ^ _ x = cos x + ce coax. 

6. j— - = sin- 1 i + c. 15 _ ^ = ce ^ 

7. ?/-m=ccosz. 16. ye mx = x n + c. 

8. ysecx=mx+c. 17. ?/e x2 = sin x + c. 

9. jSVi -a 2 = sin-^ + c. 18. y = x (ce x - 1). 

Art. 292, page 436. 

VF^ 2 " 4 - «(c-20 5 ) = 50 2 . 

1. =sin -1 z + c. 

V b. y H 2sm ~ lx = c(l -2/ 2 ). 

2. z 3 w(z + c) = 1. 

i _ r2 6. r = (2 + ccos0) 2 . 



[c - (1 - z 2 ) 1 ] 2 7. r (1 + z 2 ) = I (1 + z 2 ) 2 + c, 



■2X 



ANSWERS 475 

8 . !+/ = | (1 + x *y + c. 10. (1 - 2a: + xft e^ = ex. 

l i 2 r -i i 11% logy = xi + cx3 - 

9. - = — e* x \ e * x sinxdx + c. „ rt . 

2/ J 12. log?/ = m + ccosx. 

Art. 296, page 440. 

1. y = Cie~ ix + c 2 e hx . 6. 2/ = Ci + c 2 e 2x + c s e~ 2x . 

2. ^ = c 1 e §x + c 2 e |x . 7. 2/ = c,e x + c 2 e 2 * + c z e~ 3 x . 
Z. y = e 2x (ci& x + c 2 e-^). 8. y = Cie 4X + c 2 e x . 

4. ?/ = cie 2iz + c 2 e- 2ic , 9. 2/ = Cie x +c 2 e- a: +C3e 2a; + c 4 e~ 
y = Cl6 2x _[_ c^-2^ 10. y = ci + c 2 e* + c 3 e-* + c^ 

5. 2/ = Ci+c 2 e+ 4a \ • +c 5 e-^. 

Art. 298, page 441. 

1. ?/ = e x (Acosx + Bsinx). 2. y = c x e x + K sin (x + a) . 

3. y = Cie x + c 2 e _;r + A cos x + B sin x. 

4. y = e 2x (A cos V3 x + £ sin V3 x). 

5. 2/ = i£icos(V2x + ft) + i£ 2 cos (V2 x + /3 2 ). 

6. 2/ = C + i^sintx+a). 

Art. 300, page 444. 

I. y = e*(ci + c 2 x + C3X 2 ). 2. ?/ = e* (ci + c 2 x + c 3 x 2 ) + c 4 . 

3. y = de 2X + e x (c 2 + c 3 x) + e-*(c 4 + c&). 

4. ?/ = ce 2 * + (Ai + A 2 x) cos a; + {B x + S 2 rc) sinx. 

5. y = ci + c 2 x + c 3 x 2 + ^. cos V5 a: + 5 sin V5 a;. 

6. y = ci + c 2 x + (Ai + A 2 x) cos 3 x + (Bi + B 2 x) sin 3 x. 

7. y = e 2x (ci + c 2 x + c 3 x 2 + c 4 x 3 ). 

8. 2/ = e V2 ~ x (d + c 2 x) + e~ V2x (c 3 + c 4 x) + (Ai + A 2 x) cos V2 a; 

+ (5i + 5 2 x)sinV2x. 
Art. 302, page 447. 

1. x 1 + 2 a; + e x (A cos V3x + B sin V3~x). , 

2. 2 a: — 4 x z + a; 4 + c^-* + c 2 sin (a; + a). 

3. cie- 2x + c 2 e ix + sina:. 

>i * 1 lx( ^7 , . . Vf \ . 

4. ce _x + e- I a cos -«-as + osm -«-x 1 + xsmx. 

5. Cie - * + c^ + c 3 cos a: + c 4 sin x + 3 sin 2 x. 

6. (a + 6a:)e 2: + e x sin2a:. 

7. e x (ciCos V5x + c 2 sin V5x + sin 2 a;). 

8. Ae 2X + Be~ 2x - | (6 x 2 + 3 - 2 xe 2 x ) . 

9. K cos (2 x + /3) + i (x + x sin 2 x). 

10. (ci + c 2 x) — e 2z + — e 21 (x cos x + cos x — sin x). 

II. ci +c 2 e 2a: — ■^■e x (xsinx + cosx). 

12. cie* + c 2 e ix - e 2x (2 x 2 - 2 x + 3) - \ sin 2 x + (x + f) cos 2 x. 

13. (ci + cax) e 1 + c 3 e- 2x + | x 2 e z (3x 2 - 4x + 4) + 2 x 2 + 6x + 9. 



476 



DIFFERENTIAL AND INTEGRAL CALCULUS 



Art. 304, page 449. 

1. y = \x 2 — \ sin 2 x + ax + c 2 . 

dy 

-x + c 2 . 



</: 



Vy — sin y cos y +Ci 
3. </> = cos + a0 3 + &0 2 + c0 + a*. 

4.2= CiX 4 + C 2 X 3 + C3X 2 + C4X + c 5 . 

5. s = igt 2 + ct + c'. 

Art. 305, page 449. 

1. xe y ~ 2x = c(y -x). 

v 

2. xy = cex. 

3. 1/ sin 2 x = 2 ma; + c. 

4. cos w x = ce v ^ x . 

0. y + - — = c. 

sinj/ 

6. ?/ = cos a; + ce x . 

7. a Vl -/3 2 + |8 Vl -a 2 = c. 

8. i/ = Cie- a; +c 2 e :c +C3e- V2x +c 4 e v ' 2:E 

+ c b e~ vlx + Cee^ 31 . 



: 2 cosl -x-x+a 1 



9. 2/ = i£e 5 

+e zx sm2x, 

10. ?/ = Xe^cosf-^-x +a ) 

+ 8z 2 + 6x + i 

11. log Vt 2 + s 2 - - - tan- 1 



+ 4a;' 



6 - s = ^( Ke °'-|^'> 

7. as = Ci sin (at + c 2 ). 

8. a 2 ?/ = e ax + C1X 2 + c 2 x + c 3 . 

9. cos a; = be ax ~ v . 



13. m ViP—v 2 + v Vl-u 2 = cit. 

14. x + Vx 2 + y 2 = c. 

15. x^ = ce x . 

16. 2/ = Ae-»*sin^s+a) 

+ 2 sin x — 3 cos x. 

17. 2/ = Ae* + Be~ x + xe*. 

q 

18. y~*cos 3 x = - (sinx+cscx)+c. 

19. 2/ = (a + bx + ex 2 ) e* + x z e x . 



20. 



3 



+ tan -1 xy = c. 



12. (l+2fla; = 3y + c. 

25. y = ce 2 * + e~^(Acos^x+Bsin-^x]-3cos2x + (l-2x)sin2x. 



26. y,= Cie V2x -\- cier- 



V2X 



2 

m 



21. 2/ = Xe i sin(V2^+ a ) 

+ we' (tsint — 2cos0- 

22. (1+x 2 ) [2 - 3i/ 2 (l +x 2 )] = c?/ 2 . 

23. 4 2 = (A - x) cos 2 x + 

(B + x) sin 2 x. 

24. Vl - x 2 (y + x - x 3 ) = ex. 
V7 



e x (x sin x + x cos x — sin x + cos x). 



INDEX 

Numbers refer to pages 



Absolute value, defined, 272. 

Acceleration, angular, 112; defined, 111; 
derivative formula for, 210; resolu- 
tion of, 150. 

Agnesi (see "Witch"). 

Angle, direction, 279. 

Arc, center of mass of, 342; derivative 
of, Cartesian coordinates, 141; deriv- 
tive of, polar coordinates, 143; dif- 
ferential of, geometric interpretation, 
142; length of, 236, 250. 

Archimedes (see "Spiral"). 

Area, derivative of, Cartesian coordi- 
nates, 225; derivative of, polar coor- 
dinates, 234; expressed as a definite 
integral, 226; of a curve given by 
parametric equations, 231; of a curve 
given by a polar equation, 235; of a 
cylindrical surface, 289, 291; of a sur- 
face, expressed as a double integral, 
328; plane, center of mass of, 341; 
plane, expressed as a double integral, 
327. 

Argument, of complex number, 389; of 
function, defined, 2; real values of, 9. 

Astroid, as envelope of fines, 412; as 
hypocycloid, 131; center of curvature 
of, 156; evolute of, 158; figure of, 
131, 158; parametric equations of, 
125; x- and y- equation of, 83; exer- 
cises on, 50, 83, 125, 131, 148, 161, 234, 
238, 260, 261, 346, 415. 

Bernoulli (see "Leminscate"). 
Bernoulli's equation, 435. 
Binomial theorem, proof of, 371. 
Boyle's law for gases, 112. 
Branches of a function, 4. 

Cardioid, as epicycloid, 131; figure of, 
131; parametric equations of, 161; 
polar equation of, 138; exercises on, 
131, 138, 148, 161, 235, 238, 264, 346. 



Catenary, curve of type of, 203; figure 
of, 119; parametric equations of , 125; 
x- and y- equation of, 119; exercises 
on, 119, 125, 148, 161, 238, 262, 347. 

Cauchy's Theorem, 165. 

Caustics, 416. 

Center of curvature, 155. 

Center of mass, defined, 339; of arc of 
astroid, 343; of arc of plane curve, 
342; of curved surface, 344; of area, 
plane, 341; of area of quadrant of 
circle, 341; of surface of revolution, 
343; of portion of spherical surface, 
344; of volume of hemisphere, 339; 
of volume of octant of sphere, 340. 

Chemical reaction, speed of, 109. 

Circle, area of involute of, 252; curva- 
ture of, 154; involute of, 132; of 
curvature, 155; osculating, 155; ex- 
ercises on, 112, 161, 231, 238, 260, 347, 
411, 415, 419. 

Cissoid of Diodes, figure of, 119; para- 
metric equations of, 125; x- and y- 
equations of, 119; exercises on, 119, 
125. 

Clairaut's equation, 436. 

Cohen, Differential Equations, 421, 436, 
442, 445. 

Compound-interest law, 95. 

Computation, of e, 89; of logarithms, 
380; of log 2, 360, 381; of log 3, 381; 
of sin 20°, 354. 

Constant, absolute, 1; arbitrary, 1, 424; 
of integration, 176. 

Continuity, defined, 60, 62; of a func- 
tion of two arguments, 293. 

Convergence of series, defined, 375; 
ratio-test for, 377. 

Convexity and concavity, criterion for, 
71; defined, 70; in polar coordinates, 
135. 

Coordinates, space, 265; spherical, 335. 

Cosines, direction, 279. 



477 



478 



INDEX 



Cubical parabola, area of, 227, 230; ex- 
ercise on, 260. 

Curvature, absolute, actual, average, 
153; center of, 155; circle of, 155; 
radius of, 155; radius of in polar 
coordinates, 162. 

Curve, area under, 226; in polar coor- 
dinates, 135; length of, 236; paramet- 
ric equations of, 120; rational, 120; 
tangent to, equation of, 37, 305; tan- 
gent to, slope of, 34. 

Curves, systems of, 410. 

Curve tracing, 81, 115; examples of, 81, 
82, 115, 116, 120, 135, 137. 

Cusp, examples of, 75, 83, 117, 118, 119, 
124, 125, 126, 131. 

Cycloid, derivation of equations of, 128; 
figure of, 128; exercises on, 128, 129, 
148, 161, 233, 238, 261, 346, 418, 419. 

Cylinder, area of, 290, 291; equation of, 
268; exercises on, 291, 331, 332. 



De Moivre's Theorem, 390. 

Derivative, as ratio of velocities, 110; 
as slope of tangent, 34; as velocity, 
103; defined, 33; descriptive nota- 
tion for, 48; differential notation for 
99; logarithmic, 41, 92; mixed par- 
tial, 299, 301; of arc, Cartesian 
coordinates, 141; of arc, polar coor- 
dinates, 143; of area, Cartesian 
coordinates, 225; of area, polar coor- 
dinates, 234; of higher order, 56; 
of higher order, differential notation 
for, 101; partial, defined, 294; par- 
tial, mixed, 299; partial, of a function 
of functions, 307; partial, of higher 
order, 299; partial, of an implicit 
function, 311; total, defined, 302. 

Descartes (see "Folium"). 

Development, defined, 357; of /(a + h), 

352, 357; of cos z, 360; of e x , 360, 
374; of >, 388; of log (a + h), 359; 
of log (1 + x), 359, 370; of sin (a+ h), 

353, 358; of sin x, 353, 359, 369, 377; 
of tana;, 379; of (1 + h)*, 371, 377; 
of (1- x 2 )~h, 374. 

Differential coefficient, 98. 

Differential, defined, 97; exact, 430; 
formulae, 99; geometric representa- 
tion oT, 98; notation, remarks on, 101, 
319; of arc, formula for, 142; of arc, 
geometric interpretation of, 142; of 
higher order, 101; partial, 315; par- 
tial and total, geometric representa- 
tion of, 319; total, 315, 430. 



Differential equation, auxiliary equa- 
tion of, 438, 440, 441; Bernoulli's, 
435; Clairaut's, 436; complete primi- 
tive of, 423, 424; defined, 421; 
degree of, 422; derivation of, 424; ex- 
act, 430; geometrical interpretation 
of, 425; homogeneous, 431; linear, 
433, 438; of first order, 427; order of, 
422; ordinary, 422; partial, 422; 
particular integral of, 422; singular 
solution of, 436; solution of, 423, 426; 
special, 448; variables separable in, 
427. 

Differentiation, defined, 38; of fractions, 
44; general formula for, 113; of cir- 
cular functions, 53; of exponential 
functions, 92; of irrational functions, 
46; of implicit functions, 49, 57; of 
logarithmic functions, 90; of polyno- 
mials, 38; of trigonometric functions, 
51; partial, 293 ff; partial, remarks on, 
312; rules for, 38; successive, 56, 57. 

Diocles (see "Cissoid"). 

Direction angles, 279. 

Direction cosines, 280. 

Discontinuity, defined, 10, 60, 62; of 
f'(x), 63; of a function of two argu- 
ments, 293; removable, 22. 

Duhamel's theorem, 248. 

Ellipse, area of, 228, 231; circumference 
of, 238; equation of tangent to, 43; 
parametric equations of, 125; ring 
formed by revolution of E about an 
axis, 258; successive differentiation 
of equation of, 57; exercises on, 83, 
85, 125, 152, 161, 239, 260, 346, 410, 
415, 419. 

Ellipsoid, equation of, 271; figure of, 
272, 286; tangent plane and normal 
to, 315; volume of, 286; exercises on, 
299, 346. 

Ellipsoid of revolution, 85, 253; surface 
and volume of, 257. 

Elliptic wedge, 275, 276. 

Envelopes, 411; property of, 413. 

Epicycloid, equation of, 129; figure of, 
129; exercises on, 131, 132, 148. 

Equations, algebraic, 65; homogeneous, 
277, 386; identical, 5; parametric, 
defined, 120; parametric, derivation 
of, 126; parametric, geometrical in- 
terpretation of, 123; roots of, 65^ 

Euler's Theorem, 387. 

Evolute, defined, 157; of astroid, 157; 
of lemniscate, 126; of parabola, 157; 
properties of, 158. 



INDEX 



* 479 



Expansion of functions (see "Develop- 
ment")- 
Expansion of metal rod, 111. 
Exponent, imaginary, 388. 
Exponential curve, 161, 202. 

Factor, integrating, 429. 

Falling bodies, 205, 206, 209, 213. 

Fine, College Algebra, 408. 

Flex, defined, 72; note on term, 72; 
rule for finding, 73. 

Folium of Descartes, discussion of, 120; 
figure of, 122; differentiation of 
equation of, 58; exercise on, 233. 

Fraction, rational, defined, 8; differen- 
tiation of, 44; integration of 398 ff; 
limit of, 25. 

Fractions, partial, 400 ff. 

Function, algebraic, 8; branches of, 4; 
change of sign of , 63 ; complementary, 
444; defined, 1; depends upon con- 
stants, 3; development of, 357; dif- 
ferentiable, 163; discontinuities of, 
10; explicit and implicit defined, 3; 
expressed as a polynomial, 349; gen- 
eral theorems on, 64, 69, 70, 71; 
graph of, 9; homogeneous, defined, 
277, 386; homogeneous, Euler's theo- 
rem for, 387; hyperbolic, 392; hyper- 
'bolic, geometric interpretation of, 395; 
implicit, differentiation of, 49, 57; 
implicit, partial differentiation of, 
311; increasing and decreasing, 68; 
inverse hyperbolic, 393; irrational, 
defined, 8; irrational, differentiation 
of, 46; irrational, integration of, 193, 
196; many- valued, 4; of several 
arguments, 2, 293; rational, defined, 
8; rational, integration of, 408; 
rational, of trigonometric functions, 
127; single- valued, 4; that has no 
limit, example of, 27; transcendental, 
8; zeros of, 67. 

Functions, kinds of, 7, 8, 9; limits of, 20; 
trigonometric, infinite limits of, 26. 

Gases, Boyle's law for, 112; van der 
Waals' equation for, 113. 

Graph, breaks in, 10, 60; defined, 9; 
discontinuity of, 11, 60; of function 
of two arguments, 293; of greatest 



whole number in x, 61; of sin — 
x 

of ^, 11; of 2* 14. 

x 



27; 



Hedrick and Kellogg, Applications of 
the Calculus to Mechanics, 346. 



Hemi-sphere, center of mass of, 339. 

Horner's method, 66. 

Hyperbola, parametric equations of, 125; 

exercises on, 37, 83, 125, 161, 415, 419. 
Hyperbolic functions, 392; geometric 

properties of, 395. 
Hyperboloid, of one sheet, 273, 274; of 

two sheets, 274, 275; of revolution, 

253; exercise on, 299. 
Hypocycloid, equation of, 130; figure 

of, 130; of four cusps, 131; exercises 

on, 131, 132, 148. 

Identity, defined, 5; differs from an 
equation, 5; sign of, 5. 

Increment, defined, 32; geometric repre- 
sentation of, 32, 319; of argument, 
identical with differential, 98; par- 
tial, 294, 319; total, 302, 319. 

Indeterminate forms, 166; by Taylor'a 

theorem, 362; |j, 167; gj-, 169; OXoo, 

170; oo-oo, 172; l 00 , oo°, 0°, 172. 

Infinite point of line, 12. 

Infinite, an, defined, 19. 

Infinitesimal, difference between vari- 
able and limit, 18; substitution of 
one for another, 144, 248. 

Infinitesimals, defined, 18; Duhamel's 
Theorem of, 248; method of, 144; 
theorem of, 144, 146 

Integrand, defined, 175. 

Integral, as the limit of a sum, 240; 
calculus, fundamental theorems of, 
243, 323, 325, 332; defined, 175; 
definite, 176, 224; definite, change of 
limits in, 229; double, 322; double, 
area of surface expressed as, 328; 
double, plane area expressed as, 327; 
elliptic, 178, 239; indefinite, 176, 224; 
line, 289; multiple, 321; particular, 
of differential equation, 422; proof 
that every function has an, 246; 
tables, 180; triple, 332; triple in 
polar coordinates, 335. 

Integrating factor, 429. 

Integration, applications of in geometry 
201; applications of in kinematics 
205; by algebraic substitution, 193 
by parts, 198; by tables, 179, 190 
by trigonometric substitution, 196 
constant of, 176; defined, 175 
devices for, 181, 185; fundamental 
formulae of, 179; general remarks on, 
178; general theorems of, 176; of any 
rational function, 408; of rational 
fractions, 399 ff ; of cos TO x sin n x, 195. 



480 



INDEX 



Interval, 60. 

Involute, denned, 157; of circle, 132; of 
circle, area of, 252. 

Law of Mean, 164, 349. 

Lemniscate of Bernoulli, area of, 235; 
evolute of, 126; figure of, 135; para- 
metric equations of, 125; polar equa- 
tion of, 135; exercises on, 125, 148, 
152, 161, 233, 260, 261, 264, 346. 

Lester, The Integrals of Mechanics, 346. 

Limacon of Pascal, 138. 

Limit, a geometrical, 28; defined, 16, 
18; example of a function that has 
no, 27; difference between limit and 
value of a function, 22 ; variable may 

COS0 



sin ,1 

not have a, 19; of— —and — 



21; of 



23; of 



23; 



of ?,24 
x 



,-( 



, 1\« . arc 

1 +- ,89; of-— r 
a I chord 



140; of ^~, 146, 251. 

df{x) 

Limits, general theorems of, 28; infinite, 
of trigonometric functions, 26; of 
functions, 20. 

Limits of integration, defined, 224; 
change of, 229. 

Line in space, angle between plane and, 
283; direction cosines of, 279; equa- 
tions of, 278, 281; normal to a plane, 
282, 283; normal to a surface, 297, 
314. 

Lines in space, angle between two, 281; 
conditions that they be perpendicular 
or parallel, 280. 

Line integral, 289. 

Logarithm, base of, 90, 381; calcula- 
tions of log 2 and log 3, 381 ; Naperian 
or natural, 90, 94. 

Logarithms, computation of, 380 ; modu- 
lus of system of, 382. 

Logarithmic curve, 161. 

Maclaurin's Theorem, 349; applications 
of, 369; finite form, 360; infinite 
form, 366; for functions of any num- 
ber of variables, 382. 
See also " .Development." 

Mass (see "Center of mass"). 

Maxima and minima, criteria for, deter- 
mined by T's Theorem 363; defined, 
76, 364; determined by means of first 
derivative, 77; employment of second 
derivative in determining, 79. 



Mixed partial derivatives, defined, 299; 
general principle of, 300. 

Modulus, of a complex number, 389; of 
a system of logarithms, 382. 

Morley, 72. 

Motion, harmonic, 107, 108, 152, 207, 
211; in a resisting medium, 223; of a 
body falling to the earth from a great 
distance, 213; of a falling body, 205, 
206, 209; of projectiles, 215, 221; 
problem of, 246. 

Murray, Differential Equations, 421. 

Normal equation of a plane, 282. 
Normal to a plane, direction cosines of, 

282; equation of, 283. 
Normal to a plane curve, 37, 305. 
Normal to a surface, direction cosines 

of, 297; equations of, 297, 314. 

Osgood, Calculus, 169; Introduction to 
Infinite Series, 377. 

Pappus, theorems of, 347. 

Parabola, area of, 228; as path of pro- 
jectile, 217; center of curvature of, 
156; derivative of arc of, 148; evolute 
of, 157; exercises on, 37, 85, 135, 152, 
231, 238, 346, 411, 415, 416, 419. 

Parabolas, envelope of system of, 412; 
systems of, 201. 

Parabaloid, elliptic, equation of, 275; 
hyperbolic, equation of, 276; of revo- 
lution, 85, 219, 253, 343; volume cut 
from by plane, 287; exercises on, 289. 

Parameter, 3, 410; variable, 120; geo- 
metric interpretation of, 123. 

Parametric equations, derivation of, 
126; of curves, 120; of line in space, 
281. 

Pascal, limacon of, 138. 

Peirce, A Short Table of Integrals, 190, 
395. 

Pierpont, Theory of Functions of a Real 
Variable, 163. 

Plane in space, angle between line and, 
283; equation of, 268; intercept 
equation of, 270; normal equation of, 
282; normal to, 282; tangent to a 
surface, 296, 314. 

Point, angular, 63; double, 115; ideal, 
12; infinite, 12; isolated, 117, 118; 
maximum and minimum, 76; mul- 
tiple, 115, 127; of discontinuity, 11; 
of inflection, 72; singular, 11. 

Points, infinitely near, 28. 



INDEX 



481 



Polynomial, conditions for multiple 
root of, 362; defined, 7; factors of, 
398; of infinite degree, 374; method 
for finding real roots of, 66; quad- 
ratic, property of, 68; root of, defined, 
65; sign of, 363. 

Power series, convergent and divergent, 
375; defined, 374; ratio-test for con- 
vergence of, 377; sum of, 375. 

Primitive, complete, 423. 

Probability curve, 119. 

Projectile, equation of path of, 217; 
motion of, 215, 221; range of, 216. 

Projection, defined, 278, 279; orthog- 
onal, 278; theorems of, 278, 279. 

Quadratic factors of a polynomial, 399. 
Quadratic polynomial, property of, 68. 
Quadrature, 227, 427. 
Quadric surfaces, 271. 
Quartic surface, 276. 

Radius of curvature, 155; in polar co- 
ordinates, 162; of astroid, 156; of 
parabola, 156; theorems on, 158, 159. 



Surface, area of, as double integral, 328; 
center of mass of, 344; element of, 
330; equation of, 268; equation of 
tangent plane and of normal to, 297, 
305,315; traces of , 269 ; quadric, 271; 
quartic, 276. 

Surface of revolution, area of, 255; area 
in polar coordinates, 262; center of 
mass of, 349 ; of ring formed by revo- 
lution of ellipse, 257. 

Tangent to a curve, defined, 28; double, 
137; equation of, 37, 305; in polar 
coordinates, 134 

Tangent plane to a surface, 296, 314. 

Taylor's Theorem, 349; applications of 
the infinite form of, 369; criteria for 
maxima and minima determined by, 
363; evaluation of the indeterminate 
form §• by, 362; examples of develop- 
ment by, 358; finite form, proof of, 
355; for functions of any number of 
arguments, 382; infinite form, 366; 
remainder in, 357. 
See also "Development." 

Time- derivative, first, 106; second, 111. 



Sector, area of, 244, 249. 

Semi-cubical parabola, 75; as evolute of 
parabola, 157; length of arc of, 236; 
exercises on, 161, 230. 

Series (see "Power Series"). 

Slope of tangent, as geometric interpre- 
tation of the derivative, 34. 

Solid, center of mass of, 338; volume of, 
286. 

Solid of revolution, volume of, 253; vol- 
ume of, in polar coordinates, 262. 

Speed, 103. 

Sphere, equation of, 271; hemi-, cen- 
ter of mass of, 339; octant of, center 
of mass of, 340; surface by double 
integration, 331 ; triple integral taken 
throughout octant of, 334; volume by 
double integration, 326; exercises on, 
86, 112, 260, 291, 299, 331, 332, 346, 
347. 

Spiral, equiangular (see "Spiral, loga- 
rithmic"). 

Spiral, hyperbolic, 139; exercise on, 236. 

Spiral, logarithmic, 139; length of, 237; 
exercises on, 152, 236, 418. 

Spiral of Archimedes, 139; exercises on, 
152, 236, 238. 

Sugar, inversion of, 109, 208. 



Undetermined coefficients, method of, 
in solving differential equations, 445. 

Values, critical, 121. 

Van der Waals' equation, 113. 

Variable, defined, 1 ; dependent and in- 
dependent, 2; parameter, 120. 

Velocity, angular, 108; average or 
mean, 103; components of, 149; de- 
rivative as, 103; instantaneous, 104; 
of any change of state, 109 ; of chemi- 
cal reaction, 109; resolution of, 149; 
resolution of along a curve, 150; tan- 
gential, 109; uniform, 103. 

Volume, as a double integral, 326; as a 
triple integral, 337; cut from a para- 
boloid by a plane, 287; of ellipsoid, 
286; of ellipsoid of revolution, 257; 
of solid of revolution, 254; of solid of 
revolution, element of, 255; of solid 
of revolution in polar coordinates, 
262; of sphere by double integration, 
326; under a surface, 321. 

Wedge, elliptic, 275. 

Witch of Agnesi, x- and y- equation of, 

118; parametric equations of, 125; 

exercises on, 118, 125, 234, 262. 



JUL 3 1912 



